This deck builds logarithms from the ground up, starting from the log as the inverse of the exponential - the log IS the exponent. It covers fluent conversion between exponential and logarithmic form, evaluating by inspection, common and natural logs, domain and the vertical asymptote, and the graph as a reflection across the line y equals x. It then gives the product, quotient, and power rules with a reason for each, expanding and condensing, change of base, and the log scales: decibels, pH, and Richter. It targets four killer errors: splitting the log of a sum, pulling out only part of an exponent, ignoring a domain violation, and inverting the change-of-base fraction.
Subject: College Algebra · 146 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 17
One sentence carries this whole deck: the logarithm is the exponent. Everything else is that sentence in different clothes.
Objectives
Logs scare people because the notation looks alien. It is not. A log is just an exponent that has been given its own name.
And just as important: recognize the four moves that look legal but are not. We will meet all four head on.
Warm-up
Discussion prompt
Before we open Logarithmic Functions and Properties of Logs: without looking back, what was the main idea of Exponential Functions and Growth, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck builds exponential functions from the ground up. It contrasts a constant rate with a constant ratio in a table, explains the restrictions on the base and why they exist, and covers growth and decay graphs with their horizontal asymptote and the transformations that move it. It then introduces the natural base e as the limit of compounding more and more often, computes compound and continuous interest with real numbers, and works through growth, decay, half-life, and cooling models, fitting a model to two data points, and solving exponential equations with a common base. It targets four killer errors: confusing a variable base with a variable exponent, multiplying the base by the exponent, forgetting that a vertical shift moves the horizontal asymptote, and using the annual interest formula when the problem says monthly.
Section
Part 1
Concept
You already know how to answer this one.
\[ 2^3 = \;? \]
Now hide a different piece of the same statement. This is the question a logarithm answers.
\[ 2^{\,?} = 8 \]
In words: to what power must I raise 2 to get 8? The answer is 3, and the notation for that question is a log.
\[ \log_2 8 = 3 \]
Counterexample
Discussion prompt
In words: to what power must I raise 2 to get 8? The answer is 3, and the notation for that question is a log.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Intuition
Read the symbol like a sentence, left to right. Say base, then say gives, and the answer is the power.
\[ \underbrace{\log_{\,2}}_{\text{base } 2} \underbrace{8}_{\text{result}} = \underbrace{3}_{\text{the exponent}} \]
So whenever you see a log, the very first thing to say in your head is: this thing equals an exponent. If the answer you get is not a plausible exponent, you have made an error.
Exponentials and logs are the same relationship read from two directions - like a recipe and its shopping list. One asks for the result, the other asks for the power.
Analogy
Discussion prompt
Explain Say it out loud: the log is the exponent by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Read the symbol like a sentence, left to right. Say base, then say gives, and the answer is the power.
Picture it
Animation
Shows: A logarithm IS an exponent — a rendered Manim animation.
Rendered with Manim.
Takeaway: Read it as a question and the definition stops feeling arbitrary.
Concept
Here is the definition. The two statements below are the same statement, just written differently.
\[ \log_b x = y \quad \Longleftrightarrow \quad b^{\,y} = x \]
logarithm — The exponent you must put on the base to produce the given number. The base sits small and low in both forms; the log's answer is always the exponent.
Notice where each piece goes. The base stays the base. The log's answer becomes the exponent. The number inside the log becomes the result.
\[ \log_{\color{#a78bfa}b} {\color{#f472b6}x} = {\color{#4ade80}y} \qquad \Longleftrightarrow \qquad {\color{#a78bfa}b}^{\,{\color{#4ade80}y}} = {\color{#f472b6}x} \]
Explain it
Discussion prompt
Explain The definition, both directions to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Notice where each piece goes. The base stays the base. The log's answer becomes the exponent. The number inside the log becomes the result.
Concept
A log is the inverse of an exponential function, so it inherits that function's rules. The base has the same two restrictions it had there.
\[ b > 0, \qquad b \ne 1 \]
A base of 1 is useless: 1 raised to any power is 1, so it can never reach 8 - no exponent exists. A negative base skips values and breaks at fractional exponents.
And the number inside the log must be strictly positive, because a positive base raised to any real power is always positive.
\[ \log_b x \ \text{ is defined only when } \ x > 0 \]
argument — The number inside the logarithm - the result you are trying to reach. It must be positive, always. Zero and negatives are outside the domain.
Definition probe
Sort into buckets
Every line below is part of the definition of logarithm or of argument — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: converting both directions into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The base is the number being raised to a power, and it stays the base in the log form too.
Worked example
Write the first statement in logarithmic form, and the second in exponential form.
\[ \text{(a)}\ \ 4^3 = 64 \qquad\qquad \text{(b)}\ \ \log_2 \tfrac{1}{16} = -4 \]
For (a), find the base: it is 4
Why: The base is the number being raised to a power, and it stays the base in the log form too.
The exponent 3 becomes the log's answer, and 64 goes inside
Why: The log always equals the exponent, so 3 must land on the right side of the equals sign.
\[ 4^3 = 64 \quad \Longrightarrow \quad \log_4 64 = 3 \]
For (b), the base is 2 and the log's value is the exponent negative 4
Why: Reading the same definition backwards: whatever the log equals is what goes up on the base.
Put the argument on the other side of the equals sign
Why: The number inside the log is the result the power produces.
\[ \log_2 \tfrac{1}{16} = -4 \quad \Longrightarrow \quad 2^{-4} = \tfrac{1}{16} \]
Verify both by evaluating the exponential side
Why: 4 cubed is 4 times 4 times 4, which is 64 - statement (a) checks. And 2 to the negative 4 is 1 over 2 to the fourth, which is 1 over 16 - statement (b) checks.
\[ 4^3 = 64 \ \checkmark \qquad 2^{-4} = \frac{1}{2^4} = \frac{1}{16} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "converting both directions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: 4 cubed is 4 times 4 times 4, which is 64 - statement (a) checks. And 2 to the negative 4 is 1 over 2 to the fourth, which is 1 over 16 - statement (b) checks.
Concept
Drill this table until the conversion is automatic. Every row is one fact written twice.
| Exponential form | Logarithmic form | Said out loud |
|---|---|---|
| 3 to the 4th equals 81 | log base 3 of 81 equals 4 | 3 to the what gives 81? Four. |
| 10 to the 2nd equals 100 | log base 10 of 100 equals 2 | 10 to the what gives 100? Two. |
| 5 to the 0 equals 1 | log base 5 of 1 equals 0 | 5 to the what gives 1? Zero. |
| 2 to the negative 3 equals one eighth | log base 2 of one eighth equals negative 3 | 2 to the what gives an eighth? Negative three. |
Two facts fall straight out of that table and are worth memorizing on their own.
\[ \log_b 1 = 0 \qquad \text{and} \qquad \log_b b = 1 \]
Comparison
Comparison matrix
From The two forms, side by side: refill the Logarithmic form column from what you know. The rest of the table is as it appeared.
| Exponential form | Logarithmic form | Said out loud |
|---|---|---|
| 3 to the 4th equals 81 | log base 3 of 81 equals 4 | 3 to the what gives 81? Four. |
| 10 to the 2nd equals 100 | log base 10 of 100 equals 2 | 10 to the what gives 100? Two. |
| 5 to the 0 equals 1 | log base 5 of 1 equals 0 | 5 to the what gives 1? Zero. |
| 2 to the negative 3 equals one eighth | log base 2 of one eighth equals negative 3 | 2 to the what gives an eighth? Negative three. |
Step zero
Discussion prompt
Worked example: evaluating a log by inspection — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Turn the question into an exponent question
Answer:
Worked example
Evaluate it without a calculator.
\[ \log_5 \frac{1}{125} \]
Turn the question into an exponent question
Why: Give the unknown a name. This converts a strange symbol into a familiar equation.
\[ \log_5 \frac{1}{125} = y \quad \Longleftrightarrow \quad 5^{\,y} = \frac{1}{125} \]
Write the right side as a power of 5
Why: 125 is 5 cubed, and a reciprocal is a negative exponent - so one over 5 cubed is 5 to the negative 3.
\[ \frac{1}{125} = \frac{1}{5^3} = 5^{-3} \]
Match the exponents
Why: Same base on both sides means the exponents must be equal - that is the one-to-one property of exponentials.
\[ 5^{\,y} = 5^{-3} \ \Longrightarrow \ y = -3 \]
Verify by converting the answer back to exponential form
Why: The claim is that 5 to the negative 3 equals one over 125. Compute it: 5 cubed is 125, and the negative exponent flips it to 1 over 125. Both sides agree, so the answer stands.
\[ 5^{-3} = \frac{1}{5^3} = \frac{1}{125} \ \checkmark \qquad \boxed{\log_5 \tfrac{1}{125} = -3} \]
Picture it
Animation
Shows: Expanding and condensing — a rendered Manim animation.
Rendered with Manim.
Takeaway: Expand to solve, condense to present.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: you are asked to evaluate this, and you remember that a negative number cubed is negative.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The mistake: that uses a base of negative 2.
The fix: check the argument against the domain before you look for an exponent.
Why: The mistake: that uses a base of negative 2. The base written here is positive 2, and it may not be changed.
Trap
The tempting move: you are asked to evaluate this, and you remember that a negative number cubed is negative.
\[ \log_2(-8) = \;? \]
Answer negative 3, because negative 2 cubed is negative 8
Why: The mistake: that uses a base of negative 2. The base written here is positive 2, and it may not be changed.
\[ (-2)^3 = -8 \ \Longrightarrow \ \log_2(-8) = -3 \quad \text{\textsf{(false)}} \]
Test the claim against the definition
Why: If the answer really were negative 3, then 2 to the negative 3 would have to equal negative 8. But it equals positive one eighth. The claim collapses.
\[ 2^{-3} = \frac{1}{8} \ne -8 \]
The fix: check the argument against the domain before you look for an exponent.
\[ \log_2(-8): \quad \text{argument} = -8 \not> 0 \]
Ask what powers of 2 can produce
Why: 2 to any real power is positive: large powers give large positives, negative powers give small positives, and it never reaches zero. Negative outputs are simply not available.
\[ 2^{\,y} > 0 \quad \text{for every real } y \]
Report it as undefined
Why: No real exponent works, so the expression has no real value. Say undefined - not zero, and not negative 3.
\[ \log_2(-8) \ \text{ is undefined over the reals} \]
Notation
Annotate
From Trap: a log can never eat a negative number — read this one piece at a time. What is each part doing?
On: \( (-2)^3 = -8 \ \Longrightarrow \ \log_2(-8) = -3 \quad \text{\textsf{(false)}} \)
Estimation
Predict first
Logs are not always whole numbers. Evaluate this one.
Commit before you compute: what does Worked example: when the answer is a fraction come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by raising the base to the answer
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. 16 to the three quarters means the fourth root of 16, cubed.
Worked example
Logs are not always whole numbers. Evaluate this one.
\[ \log_{16} 8 \]
Notice 16 is smaller-based than it looks: both 16 and 8 are powers of 2
Why: When the base and the argument are not powers of each other, look for a common smaller base. Here 16 is 2 to the fourth and 8 is 2 cubed.
\[ 16 = 2^4, \qquad 8 = 2^3 \]
Set up the exponent equation
Why: Name the unknown exponent, then rewrite both sides with base 2 so the exponents can be compared.
\[ 16^{\,y} = 8 \ \Longrightarrow \ (2^4)^{\,y} = 2^3 \ \Longrightarrow \ 2^{4y} = 2^3 \]
Match exponents and solve
Why: Equal bases force equal exponents, leaving a one-step linear equation.
\[ 4y = 3 \ \Longrightarrow \ y = \frac{3}{4} \]
Sanity-check the size before verifying
Why: 8 is less than 16 but more than 1, so the exponent must sit strictly between 0 and 1. Three quarters passes that smell test.
Verify by raising the base to the answer
Why: 16 to the three quarters means the fourth root of 16, cubed. The fourth root of 16 is 2, and 2 cubed is 8 - which is exactly the argument. Confirmed.
\[ 16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8 \ \checkmark \qquad \boxed{\log_{16} 8 = \tfrac{3}{4}} \]
Picture it
Animation
Shows: Each line of the worked example "when the answer is a fraction", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: 16 to the three quarters means the fourth root of 16, cubed. The fourth root of 16 is 2, and 2 cubed is 8 - which is exactly the argument. Confirmed.
Ranking
Put in order
These are the steps of Pattern: evaluating any log by inspection, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Four steps. They never change, no matter how ugly the numbers look.
Then verify: raise the base to the answer you got and confirm you land exactly on the argument.
| If the argument is... | Expect an answer that is... |
|---|---|
| bigger than the base | greater than 1 |
| equal to the base | exactly 1 |
| between 1 and the base | a fraction between 0 and 1 |
| exactly 1 | zero |
| between 0 and 1 | negative |
Trade off
Comparison matrix
From Pattern: evaluating any log by inspection: every row here is a choice with a cost. Fill the Expect an answer that is... column, then say which row you would actually pick and what you give up for it.
| If the argument is... | Expect an answer that is... |
|---|---|
| bigger than the base | greater than 1 |
| equal to the base | exactly 1 |
| between 1 and the base | a fraction between 0 and 1 |
| exactly 1 | zero |
| between 0 and 1 | negative |
Elimination
Eliminate the wrong options
What is the value of the base-4 logarithm of 64?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Ask: 4 to what power gives 64? Since 4 times 4 is 16 and 16 times 4 is 64, the exponent is 3. Verify by converting back: 4 cubed equals 64, so the answer is 3.
Check
Convert it to an exponent question first, then answer.
\[ \log_4 64 = \;? \]
Check your understanding
What is the value of the base-4 logarithm of 64?
Answer: A
Why: Ask: 4 to what power gives 64? Since 4 times 4 is 16 and 16 times 4 is 64, the exponent is 3. Verify by converting back: 4 cubed equals 64, so the answer is 3.
Section
Part 2
Concept
Two bases show up so often that they were given shorthand. The first is 10 - the base our whole number system already runs on.
When a log is written with no base at all, the base is 10. That is the only time you may assume a base.
\[ \log x \quad \text{means} \quad \log_{10} x \]
common logarithm — A logarithm with base 10, written without showing the base. It answers: 10 to what power gives this number? It is the LOG button on every calculator.
This is the log behind decibels, pH, and the Richter scale - all of which we compute at the end of this deck.
Picture it
Animation
Shows: The property that does not exist — a rendered Manim animation.
Rendered with Manim.
Takeaway: The single most common logarithm error.
Missing information
Discussion prompt
No calculator. Every one of these is a power of 10 in disguise.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
1000 is 10 times 10 times 10, so it is 10 raised to the third power. The log is that exponent.
Worked example
No calculator. Every one of these is a power of 10 in disguise.
\[ \text{(a)}\ \log 1000 \qquad \text{(b)}\ \log 0.001 \qquad \text{(c)}\ \log 1 \]
For (a), count the zeros
Why: 1000 is 10 times 10 times 10, so it is 10 raised to the third power. The log is that exponent.
\[ 1000 = 10^3 \ \Longrightarrow \ \log 1000 = 3 \]
For (b), rewrite the decimal as a fraction first
Why: 0.001 is one thousandth, and a reciprocal power of 10 carries a negative exponent. Decimals hide their exponent; fractions show it.
\[ 0.001 = \frac{1}{1000} = \frac{1}{10^3} = 10^{-3} \ \Longrightarrow \ \log 0.001 = -3 \]
For (c), remember what makes any base equal 1
Why: Only the zero exponent produces 1, no matter what the base is. So the log of 1 is zero in every base.
\[ 10^0 = 1 \ \Longrightarrow \ \log 1 = 0 \]
Verify all three by raising 10 to each answer
Why: 10 cubed is 1000, matching (a). 10 to the negative 3 is 1 over 1000, which is 0.001, matching (b). 10 to the zero is 1, matching (c). All three land exactly on their arguments.
\[ 10^{3} = 1000\ \checkmark \qquad 10^{-3} = 0.001\ \checkmark \qquad 10^{0} = 1\ \checkmark \]
Concept
The second shorthand base is the number that governs continuous growth. It has its own letter and its own log symbol.
\[ e \approx 2.718281828\ldots \]
A log with base e is called the natural log and gets its own three-letter name. The base is never written.
\[ \ln x \quad \text{means} \quad \log_e x \]
natural logarithm — The logarithm with base e, written ln. It answers: e to what power gives this number? Every rule for logs applies to it unchanged - it is not a special case, just a named base.
Do not let the name intimidate you. Anywhere you would write a base, ln simply has e sitting there quietly.
\[ \ln e = 1, \qquad \ln 1 = 0 \]
Picture it
Animation
Shows: Log and exponential are mirror images — a rendered Manim animation.
Rendered with Manim.
Takeaway: Reflected across y = x, because each undoes the other.
Intuition
Picture money in an account that grows continuously - not once a year, not once a month, but every instant, at 100 percent per year.
After one year, one dollar becomes e dollars. After two years, e times e. The base e is what continuous growth does to 1 in one time unit.
So the natural log answers a very physical question: how long must that continuous growth run to reach a given multiple?
\[ \ln 7.389\ldots = 2 \quad \text{means: two time units of continuous growth} \]
That is why the natural log turns up in half-life, cooling, and interest problems - it is the clock of continuous change.
Picture it
Animation
Shows: Why ln gets its own name — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every other base carries a stray constant. Base e does not.
Worked example
The base is e. Ask the same question you always ask: e to what power?
\[ \text{(a)}\ \ln e^{4} \qquad \text{(b)}\ \ln \frac{1}{e} \qquad \text{(c)}\ \ln \sqrt{e} \]
For (a), read the exponent straight off
Why: The argument is already written as e raised to a power, and the log is that power. Nothing to compute.
\[ \ln e^{4} = 4 \]
For (b), turn the reciprocal into a negative exponent
Why: One over a base is that base to the negative first power. Once the argument is a power of e, the log is that exponent.
\[ \frac{1}{e} = e^{-1} \ \Longrightarrow \ \ln \frac{1}{e} = -1 \]
For (c), turn the radical into a rational exponent
Why: A square root is the one-half power. Radicals are exponents wearing a hat, and logs only care about exponents.
\[ \sqrt{e} = e^{1/2} \ \Longrightarrow \ \ln \sqrt{e} = \frac{1}{2} \]
Verify by raising e to each answer and comparing with the argument
Why: e to the fourth is the argument in (a). e to the negative 1 is 1 over e, the argument in (b). e to the one-half is the square root of e, the argument in (c). Every one returns exactly what was inside the log.
\[ e^{4} = e^{4}\ \checkmark \qquad e^{-1} = \tfrac{1}{e}\ \checkmark \qquad e^{1/2} = \sqrt{e}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "natural logs by inspection", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: e to the fourth is the argument in (a). e to the negative 1 is 1 over e, the argument in (b). e to the one-half is the square root of e, the argument in (c). Every one returns exactly what was inside the log.
Concept
Because a log is the inverse of an exponential with the same base, doing one and then the other returns you to where you started.
\[ \log_b\!\left(b^{\,x}\right) = x \qquad \text{and} \qquad b^{\,\log_b x} = x \ \ (x > 0) \]
Read the first one aloud: the exponent you must put on the base to get the base raised to that exponent - it is obviously that exponent.
Read the second: the log is by definition the power that turns the base into the argument. So putting it back on the base rebuilds the argument.
Think of a lock and its key, or an elevator up and the same elevator down. These two identities do more work than any other fact in this deck.
Fill the middle
Fill in the blanks
From Worked example: simplifying with the round trip — finish the line. Write what belongs on the right of the equals sign before you look.
\log_7\!\left(7^2x + 1\right) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Both are 3, so the log and the exponential cancel and only the exponent survives - negative sign included.
Worked example
Simplify each without a calculator.
\[ \text{(a)}\ \log_3\!\left(3^{-5}\right) \qquad \text{(b)}\ e^{\ln 12} \qquad \text{(c)}\ \log_7\!\left(7^{\,2x+1}\right) \]
For (a), match the base of the log to the base of the power
Why: Both are 3, so the log and the exponential cancel and only the exponent survives - negative sign included.
\[ \log_3\!\left(3^{-5}\right) = -5 \]
For (b), notice the exponential is on the outside this time
Why: The natural log has base e and so does the outer exponential, so the round trip runs the other direction and hands back the argument.
\[ e^{\ln 12} = 12 \]
For (c), let the exponent be a whole expression
Why: The identity does not care whether the exponent is a number or an expression - the log returns whatever sat up there, in one piece.
\[ \log_7\!\left(7^{\,2x+1}\right) = 2x + 1 \]
Verify each by rebuilding the original expression
Why: Put negative 5 back on base 3 and you get 3 to the negative 5, the original argument in (a). Take the log base e of 12 and put it on e and you get 12 back in (b). Put 2x plus 1 back on base 7 and you recover the original power in (c). Every round trip closes.
\[ 3^{-5} \ \checkmark \qquad \ln 12 \ \text{on } e \to 12 \ \checkmark \qquad 7^{\,2x+1} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "simplifying with the round trip", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Put negative 5 back on base 3 and you get 3 to the negative 5, the original argument in (a). Take the log base e of 12 and put it on e and you get 12 back in (b). Put 2x plus 1 back on base 7 and you recover the original power in (c). Every round trip closes.
Prediction
Predict first
What is the natural log of 1 divided by e cubed?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: negative 3
Why: One over e cubed is e to the negative 3, and the natural log returns that exponent, so the value is negative 3. Verify by converting back: e to the negative 3 is 1 over e cubed, exactly the argument.
Check
Rewrite the argument as a power of e first, then read off the exponent.
\[ \ln \frac{1}{e^{3}} = \;? \]
Check your understanding
What is the natural log of 1 divided by e cubed?
Answer: A
Why: One over e cubed is e to the negative 3, and the natural log returns that exponent, so the value is negative 3. Verify by converting back: e to the negative 3 is 1 over e cubed, exactly the argument.
Concept
We know the argument must be positive. When the argument contains a variable, that sentence becomes an inequality you actually have to solve.
\[ f(x) = \log_b\big(\text{argument}\big) \quad \Longrightarrow \quad \text{argument} > 0 \]
That is the whole method. Set the inside greater than zero, solve, and write the answer in interval notation.
Strictly greater than - never greater than or equal to. Zero is excluded, because no power of a positive base ever reaches zero.
Picture it
Animation
Shows: You can only take a log of a positive — a rendered Manim animation.
Rendered with Manim.
Takeaway: The domain restriction is where extraneous solutions get caught later.
Step zero
Discussion prompt
Worked example: finding a domain — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Set the entire argument greater than zero
Answer:
Worked example
Find the domain and write it in interval notation.
\[ f(x) = \log_5\left(2x - 6\right) \]
Set the entire argument greater than zero
Why: The argument is everything inside the log. A base-5 power can only produce positive outputs, so only positive inputs are reachable.
\[ 2x - 6 > 0 \]
Add 6 to both sides
Why: Undo the subtraction to isolate the variable term. Adding the same amount to both sides never changes an inequality's direction.
\[ 2x > 6 \]
Divide both sides by 2
Why: 2 is positive, so the inequality symbol stays pointing the same way. Only dividing by a negative would flip it.
\[ x > 3 \]
Write the answer in interval notation
Why: An open parenthesis at 3 because 3 itself is excluded, and infinity always takes a parenthesis.
\[ \text{Domain: } (3, \infty) \]
Verify with one point inside and one point outside
Why: At x equals 4 the argument is 2 times 4 minus 6, which is 2 - positive, so the log is defined. At x equals 2 the argument is negative 2 - not allowed. And at x equals 3 the argument is exactly 0, also not allowed. The boundary lands right where we said.
\[ f(4) = \log_5 2 \ \checkmark \qquad f(3) = \log_5 0 \ \text{undefined} \qquad f(2) = \log_5(-2) \ \text{undefined} \]
Picture it
Animation
Shows: Each line of the worked example "finding a domain", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equals 4 the argument is 2 times 4 minus 6, which is 2 - positive, so the log is defined. At x equals 2 the argument is negative 2 - not allowed. And at x equals 3 the argument is exactly 0, also not allowed. The boundary lands right where we said.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: you remember the rule as the log needs a positive number, so you make the variable positive.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The mistake: the rule is about the argument, not about the letter x.
The fix: put the whole argument into the inequality, then solve it.
Why: The mistake: the rule is about the argument, not about the letter x. Here the argument is x minus 5, not x.
Trap
The tempting move: you remember the rule as the log needs a positive number, so you make the variable positive.
\[ f(x) = \log_2\left(x - 5\right) \]
Declare the domain to be all positive x
Why: The mistake: the rule is about the argument, not about the letter x. Here the argument is x minus 5, not x.
\[ x > 0 \ \Longrightarrow \ \text{Domain} = (0, \infty) \quad \text{\textsf{(false)}} \]
Test a point that this claim allows
Why: x equals 1 is inside the claimed domain, but the argument becomes 1 minus 5, which is negative 4. The function is undefined there, so the claim is wrong.
\[ f(1) = \log_2(1 - 5) = \log_2(-4) \ \text{undefined} \]
The fix: put the whole argument into the inequality, then solve it.
\[ x - 5 > 0 \]
Solve the inequality for x
Why: Add 5 to both sides. The boundary is wherever the argument would hit zero, and that is what shifts the domain away from the origin.
\[ x > 5 \ \Longrightarrow \ \text{Domain} = (5, \infty) \]
Test the same two points against the corrected domain
Why: x equals 1 is now correctly excluded. x equals 6 gives an argument of 1, and the log of 1 is 0 - defined, as promised. The corrected interval keeps exactly the legal inputs.
\[ f(6) = \log_2 1 = 0 \ \checkmark \qquad x = 1 \ \text{excluded} \ \checkmark \]
Concept
As the argument creeps down toward zero, the exponent needed to produce it runs away downward without limit.
\[ 2^{-10} = \tfrac{1}{1024}, \quad 2^{-20} = \tfrac{1}{1048576}, \quad \ldots \]
Tiny positive arguments demand huge negative logs. The graph plunges but never touches the line where the argument equals zero.
vertical asymptote — A vertical line the graph approaches but never reaches. For a log function it sits exactly where the argument equals zero - the boundary of the domain.
So finding the asymptote is free once you have the domain: it is the boundary number, written as a vertical line.
\[ f(x) = \log_5(2x - 6): \ \text{domain } (3,\infty), \ \text{asymptote } x = 3 \]
Missing information
Discussion prompt
Find the domain and the vertical asymptote. Watch the sign in front of the variable.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Same rule as always - the natural log is just base e, and it obeys the same domain requirement.
Worked example
Find the domain and the vertical asymptote. Watch the sign in front of the variable.
\[ g(x) = \ln\left(4 - x\right) \]
Set the argument greater than zero
Why: Same rule as always - the natural log is just base e, and it obeys the same domain requirement.
\[ 4 - x > 0 \]
Add x to both sides instead of subtracting 4
Why: Moving the variable to the positive side avoids dividing by a negative later, which is where the flip-the-symbol mistakes happen.
\[ 4 > x \]
Read it with the variable first
Why: The statement 4 is greater than x says exactly the same thing as x is less than 4. Rewriting it this way makes the interval obvious.
\[ x < 4 \ \Longrightarrow \ \text{Domain} = (-\infty, 4) \]
Name the asymptote at the boundary
Why: The argument is zero when x equals 4, so that vertical line is the wall the graph runs alongside. Here the graph lives to the LEFT of the wall, because the domain does.
\[ \text{Vertical asymptote: } x = 4 \]
Verify with a test point on each side of 4
Why: At x equals 0 the argument is 4, so the value is the natural log of 4 - defined. At x equals 5 the argument is negative 1 - undefined. At x equals 3 the argument is 1, giving a value of 0. The domain is exactly the numbers below 4.
\[ g(0) = \ln 4\ \checkmark \qquad g(3) = \ln 1 = 0\ \checkmark \qquad g(5) = \ln(-1)\ \text{undefined} \]
Picture it
Animation
Shows: Each line of the worked example "domain and asymptote when x is subtracted", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equals 0 the argument is 4, so the value is the natural log of 4 - defined. At x equals 5 the argument is negative 1 - undefined. At x equals 3 the argument is 1, giving a value of 0. The domain is exactly the numbers below 4.
Check
Set the argument greater than zero and solve before you look at the choices.
\[ f(x) = \log_3\left(5 - x\right) \]
Check your understanding
What is the domain of this function in interval notation?
Answer: A
Why: Set the argument 5 minus x greater than zero, add x to both sides to get 5 greater than x, so x is less than 5. Verify at x equals 4: the argument is 1 and the log of 1 is 0, which is defined; at x equals 6 the argument is negative 1, which is not.
Section
Part 3
Picture it
Figure (svg): An exponential curve rising steeply, a logarithmic curve rising slowly, and a dashed diagonal line between them showing that each curve is the mirror image of the other.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Inverse functions swap inputs and outputs. Graphically, that swap is a reflection across the diagonal line where the output equals the input.
Concept
Inverse functions swap inputs and outputs. Graphically, that swap is a reflection across the diagonal line where the output equals the input.
\[ y = x \]
Figure (svg): An exponential curve rising steeply, a logarithmic curve rising slowly, and a dashed diagonal line between them showing that each curve is the mirror image of the other.
Every feature swaps with it. The exponential's horizontal asymptote becomes the log's vertical asymptote. Its y-intercept becomes the log's x-intercept.
| Exponential | Logarithm |
|---|---|
| domain: all real numbers | range: all real numbers |
| range: positive numbers only | domain: positive numbers only |
| horizontal asymptote at the x-axis | vertical asymptote at the y-axis |
| passes through the point 0 comma 1 | passes through the point 1 comma 0 |
Comparison
Comparison matrix
From The log graph is the exponential graph, reflected: refill the Logarithm column from what you know. The rest of the table is as it appeared.
| Exponential | Logarithm |
|---|---|
| domain: all real numbers | range: all real numbers |
| range: positive numbers only | domain: positive numbers only |
| horizontal asymptote at the x-axis | vertical asymptote at the y-axis |
| passes through the point 0 comma 1 | passes through the point 1 comma 0 |
Picture it
Animation
Shows: The logarithm climbs ever more slowly — a rendered Manim animation.
Rendered with Manim.
Takeaway: Vertical asymptote at zero, and no ceiling at all.
Concept
You never need a table of ten values. Two points and the asymptote determine the whole shape.
\[ y = \log_b x \quad \text{passes through} \quad (1, 0) \ \text{ and } \ (b, 1) \]
The first point is the identity that the log of 1 is zero. The second is the identity that the log of the base is one. Both are free.
Add the wall on the left and the direction of travel, and the sketch is done.
Estimation
Predict first
Sketch it by finding points, the asymptote, and the direction.
Commit before you compute: what does Worked example: sketching a base-3 log come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify every point by converting it back to exponential form
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The point 9 comma 2 claims 3 squared equals 9 - true.
Worked example
Sketch it by finding points, the asymptote, and the direction.
\[ y = \log_3 x \]
Choose x-values that are powers of 3
Why: Powers of the base are the only inputs whose logs you can name exactly. Picking anything else forces a calculator for no reason.
\[ x = \tfrac{1}{9},\ \tfrac{1}{3},\ 1,\ 3,\ 9 \]
Read each output as the exponent on 3
Why: 3 to the negative 2 is one ninth, 3 to the negative 1 is one third, 3 to the zero is 1, 3 to the first is 3, and 3 squared is 9. The log returns those exponents.
| x | y equals log base 3 of x |
|---|---|
| one ninth | negative 2 |
| one third | negative 1 |
| 1 | 0 |
| 3 | 1 |
| 9 | 2 |
Draw the vertical asymptote at the y-axis
Why: The argument is just x, so it hits zero at x equals 0. The curve hugs that line going downward and never crosses it.
\[ x = 0 \]
Connect the points as a slowly rising curve
Why: Notice the x-values triple while the y-values only step up by 1 each time. That is what makes a log flatten out - it needs three times the input for one more unit of output.
Verify every point by converting it back to exponential form
Why: The point 9 comma 2 claims 3 squared equals 9 - true. The point 1 comma 0 claims 3 to the zero equals 1 - true. The point one ninth comma negative 2 claims 3 to the negative 2 equals one ninth - true, since 3 squared is 9 and the negative flips it. Every plotted point is a true exponential statement.
\[ 3^{2} = 9\ \checkmark \qquad 3^{0} = 1\ \checkmark \qquad 3^{-2} = \tfrac{1}{9}\ \checkmark \]
Picture it
Animation
Shows: Two values worth knowing instantly — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both immediate from the definition — no work required.
Concept
Changes inside the log act on the input, so they move the domain - and the asymptote moves with it.
\[ y = \log_b(x - h) \ \Longrightarrow \ \text{asymptote at } x = h \]
As always with horizontal shifts, the graph moves opposite to the sign you read. Subtracting inside shifts right; adding inside shifts left.
Changes outside the log act on the output. They slide or stretch the curve up and down, but the wall stays exactly where it was.
\[ y = \log_b(x) + k \ \Longrightarrow \ \text{asymptote still } x = 0 \]
| Change | Effect on the graph | Asymptote |
|---|---|---|
| subtract h inside | shift right by h | moves to x equals h |
| add k outside | shift up by k | unchanged |
| multiply the log by a | vertical stretch by a | unchanged |
| negate the whole log | reflect across the x-axis | unchanged |
| negate the argument | reflect across the y-axis | unchanged, but the domain flips side |
Discrimination
Sort into buckets
Sort these by Asymptote, from memory, without looking back at Transformations: only the inside moves the wall. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Ranking
Put in order
Put the moves of Worked example: a shifted log graph into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Adding 3 inside is a horizontal shift; subtracting 1 outside is a vertical shift.
Worked example
Give the domain, the vertical asymptote, and two points on the graph.
\[ y = \log_2(x + 3) - 1 \]
Separate the inside change from the outside change
Why: Adding 3 inside is a horizontal shift; subtracting 1 outside is a vertical shift. Only the inside one can affect the domain.
\[ \text{inside: } +3 \qquad \text{outside: } -1 \]
Solve the argument greater than zero for the domain
Why: Adding 3 inside shifts the whole picture 3 units to the left, and the domain boundary moves with it.
\[ x + 3 > 0 \ \Longrightarrow \ x > -3 \ \Longrightarrow \ (-3, \infty) \]
Name the asymptote at the boundary
Why: The argument equals zero at negative 3, so that vertical line is the wall. The vertical shift of 1 does nothing to it.
\[ x = -3 \]
Pick x-values that make the argument a power of 2
Why: Choose x so that x plus 3 is 1, 4, or 8. Those give exact logs and no calculator.
| x | argument x plus 3 | log base 2 | y after subtracting 1 |
|---|---|---|---|
| negative 2 | 1 | 0 | negative 1 |
| 1 | 4 | 2 | 1 |
| 5 | 8 | 3 | 2 |
Verify the point 1 comma 1 in the original equation
Why: Substitute x equals 1: the argument becomes 1 plus 3, which is 4. The base-2 log of 4 is 2 because 2 squared is 4. Then subtract 1 to get 1, which matches the y-value claimed. The point checks.
\[ \log_2(1 + 3) - 1 = \log_2 4 - 1 = 2 - 1 = 1 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a shifted log graph", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substitute x equals 1: the argument becomes 1 plus 3, which is 4. The base-2 log of 4 is 2 because 2 squared is 4. Then subtract 1 to get 1, which matches the y-value claimed. The point checks.
Commit first
Predict first
Which single change moves the vertical asymptote away from the y-axis?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: replacing the argument x with x minus 4
Why: The asymptote sits where the argument equals zero. Changing the argument to x minus 4 makes it zero at x equals 4, so the domain becomes all numbers greater than 4 and the wall moves to that line. The other three changes only rescale or slide outputs.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Start from the plain base-2 log, whose vertical asymptote is the y-axis.
\[ y = \log_2 x \quad \text{has asymptote} \quad x = 0 \]
Check your understanding
Which single change moves the vertical asymptote away from the y-axis?
Answer: A
Why: The asymptote sits where the argument equals zero. Changing the argument to x minus 4 makes it zero at x equals 4, so the domain becomes all numbers greater than 4 and the wall moves to that line. The other three changes only rescale or slide outputs.
Section
Part 4
Concept
Here is the first of the three rules that make logs useful rather than just curious.
\[ \log_b(MN) = \log_b M + \log_b N \qquad (M > 0,\ N > 0) \]
The one-line reason: logs are exponents, and multiplying powers of the same base adds their exponents. The rule is that exponent law read backwards.
\[ b^{\,m} \cdot b^{\,n} = b^{\,m+n} \]
Notice the shape of the trade: the heavy operation on the inside becomes the light operation on the outside. Multiplication in, addition out.
Picture it
Animation
Shows: Products become sums — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every log property is an exponent property, read backwards.
Intuition
Before calculators, multiplying two six-digit numbers by hand took minutes and invited errors. Adding two six-digit numbers took seconds.
So astronomers looked up the logs of both numbers, added them, and looked the answer back up. Centuries of navigation and astronomy ran on that trick.
That is literally what a slide rule is: two log scales sliding past each other, so that adding lengths multiplies numbers.
Keep that image. Every log property is the same trade - it demotes an operation one level. Multiply becomes add, divide becomes subtract, and a power becomes a plain multiplier.
Hypothesis
Predict first
Worked example: the product rule, checked with real numbers is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Compute the left side directly
Why: Multiply inside first, then ask the log question. 4 times 8 is 32, and 32 is 2 to the fifth power.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Do not take the rule on faith. Test it where every value is exact.
\[ \log_2(4 \cdot 8) \]
Compute the left side directly
Why: Multiply inside first, then ask the log question. 4 times 8 is 32, and 32 is 2 to the fifth power.
\[ \log_2(4 \cdot 8) = \log_2 32 = 5 \]
Now split it with the product rule
Why: The rule promises the same value from two separate logs added together. Both arguments here are exact powers of 2.
\[ \log_2 4 + \log_2 8 \]
Evaluate each piece
Why: 2 squared is 4, so the first log is 2. 2 cubed is 8, so the second is 3.
\[ \log_2 4 + \log_2 8 = 2 + 3 = 5 \]
See the exponent law underneath
Why: The exponents 2 and 3 added to 5 for exactly the same reason that 4 times 8 equals 32 - the powers of 2 combined.
\[ 2^2 \cdot 2^3 = 2^{2+3} = 2^5 = 32 \]
Verify by converting the shared answer back to exponential form
Why: The claim is that the base-2 log of 32 is 5. Raise 2 to the fifth: 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32. It lands exactly on the argument, and both routes gave the same 5.
\[ 2^5 = 32 \ \checkmark \qquad \log_2(4\cdot 8) = \log_2 4 + \log_2 8 = 5 \]
Picture it
Animation
Shows: Each line of the worked example "the product rule, checked with real numbers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The claim is that the base-2 log of 32 is 5. Raise 2 to the fifth: 2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32. It lands exactly on the argument, and both routes gave the same 5.
Fill the middle
Fill in the blanks
From Trap: the log of a SUM does not split — finish the line. Write what belongs on the right of the equals sign before you look.
\log_2(8 + 8) = \log_2 16
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The mistake: the product rule applies to a product only.
Trap
The tempting move: the product rule felt like a distribution, so you distribute the log across a plus sign too.
\[ \log_2(8 + 8) \;\overset{?}{=}\; \log_2 8 + \log_2 8 \]
Split the sum and evaluate each log
Why: The mistake: the product rule applies to a product only. There is no rule at all for the log of a sum.
\[ \log_2 8 + \log_2 8 = 3 + 3 = 6 \quad \text{\textsf{(false)}} \]
Test the claim against the definition
Why: If the answer really were 6, then 2 to the sixth would have to equal 16. But 2 to the sixth is 64, four times too big. The claim is off by a factor of 4.
\[ 2^6 = 64 \ne 16 \]
The fix: add inside the parentheses first, then take one log of the single number you get.
\[ \log_2(8 + 8) = \log_2 16 \]
Evaluate the single log
Why: 16 is 2 to the fourth power, so the log is 4. One argument, one log, one answer.
\[ \log_2 16 = 4 \]
Verify by converting back
Why: 2 to the fourth is 16, which is exactly the argument after the addition. The honest value is 4, not the 6 the illegal split produced. Remember the slogan: logs break up products, never sums.
\[ 2^4 = 16 \ \checkmark \qquad \log_b(M+N) \ne \log_b M + \log_b N \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The fix: add inside the parentheses first, then take one log of the single number you get.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
The mistake: the product rule applies to a product only. There is no rule at all for the log of a sum.
Concept
Same trade, one level down on the other side.
\[ \log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N \qquad (M > 0,\ N > 0) \]
The one-line reason: dividing powers of the same base subtracts their exponents, and logs are exponents.
\[ \frac{b^{\,m}}{b^{\,n}} = b^{\,m-n} \]
Order matters here in a way it did not for the product rule. The numerator's log comes first, and the denominator's is what gets subtracted.
Step zero
Discussion prompt
Worked example: the quotient rule, checked with real numbers — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Compute the left side directly
Answer:
Worked example
Again, test it where every log is exact.
\[ \log_3\!\left(\frac{81}{9}\right) \]
Compute the left side directly
Why: Divide inside first: 81 divided by 9 is 9, and 9 is 3 squared.
\[ \log_3\!\left(\frac{81}{9}\right) = \log_3 9 = 2 \]
Now split it with the quotient rule
Why: Numerator log minus denominator log, in that order.
\[ \log_3 81 - \log_3 9 \]
Evaluate each piece
Why: 3 to the fourth is 81, so the first log is 4. 3 squared is 9, so the second is 2.
\[ \log_3 81 - \log_3 9 = 4 - 2 = 2 \]
Notice what reversing the order would cost
Why: Subtracting in the wrong order gives 2 minus 4, which is negative 2 - the reciprocal answer. That would be the log of 9 over 81, a different problem entirely.
\[ \log_3 9 - \log_3 81 = -2 = \log_3\!\left(\tfrac{9}{81}\right) \]
Verify by converting the answer back to exponential form
Why: The claim is that the base-3 log of 9 is 2. Raise 3 to the second power: 3 times 3 is 9, which is exactly 81 divided by 9. Both routes agree on 2.
\[ 3^2 = 9 = \frac{81}{9} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the quotient rule, checked with real numbers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The claim is that the base-3 log of 9 is 2. Raise 3 to the second power: 3 times 3 is 9, which is exactly 81 divided by 9. Both routes agree on 2.
Concept
The third rule is the workhorse. It is what lets logs solve equations where the unknown is stuck up in an exponent.
\[ \log_b\!\left(M^{\,p}\right) = p \cdot \log_b M \qquad (M > 0) \]
The one-line reason: raising a power to a power multiplies the exponents, so the outer exponent becomes a plain multiplier out front.
\[ \left(b^{\,m}\right)^{p} = b^{\,mp} \]
You can also read it as repeated use of the product rule: the log of M times M times M is three copies of the log of M added up, which is three times it.
Radicals count as powers here. A square root is the one-half power, so its log carries a one-half out front.
\[ \log_b \sqrt{M} = \log_b M^{1/2} = \tfrac{1}{2}\log_b M \]
Fill the middle
Fill in the blanks
From Worked example: the power rule, checked with real numbers — finish the line. Write what belongs on the right of the equals sign before you look.
\log_2\!\left(8^4\log_2 8\right) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The exponent 4 sits on the entire argument, so it is allowed to become a multiplier.
Worked example
One more exact test before we trust it on variables.
\[ \log_2\!\left(8^{4}\right) \]
Apply the power rule to move the 4 out front
Why: The exponent 4 sits on the entire argument, so it is allowed to become a multiplier.
\[ \log_2\!\left(8^{4}\right) = 4\log_2 8 \]
Evaluate the leftover log and multiply
Why: 2 cubed is 8, so the base-2 log of 8 is 3. Then 4 times 3 is 12.
\[ 4\log_2 8 = 4 \cdot 3 = 12 \]
Compute the argument the long way as a cross-check
Why: 8 to the fourth is 8 times 8 times 8 times 8. That is 64 times 64, which is 4096.
\[ 8^4 = 64 \cdot 64 = 4096 \]
Express 4096 as a power of 2
Why: 8 is 2 cubed, so 8 to the fourth is 2 to the twelfth. Doubling from 2 to the tenth, which is 1024, gives 2048 and then 4096 - so 4096 really is 2 to the twelfth.
\[ 8^4 = \left(2^3\right)^4 = 2^{12} = 4096 \]
Verify by converting the answer back to exponential form
Why: The claim is that the base-2 log of 4096 is 12, which says 2 to the twelfth equals 4096. We just built 4096 as 2 to the twelfth by hand. The power rule and the direct computation agree exactly.
\[ 2^{12} = 4096 = 8^4 \ \checkmark \qquad \boxed{\log_2\!\left(8^4\right) = 12} \]
Picture it
Animation
Shows: Each line of the worked example "the power rule, checked with real numbers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The claim is that the base-2 log of 4096 is 12, which says 2 to the twelfth equals 4096. We just built 4096 as 2 to the twelfth by hand. The power rule and the direct computation agree exactly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: you see an exponent anywhere inside and drag it out front.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The mistake: the 2 is an exponent on x alone, not on the product 9 times x.
The fix: split the product first with the product rule, then apply the power rule only to the factor that actually carries the exponent.
Why: The mistake: the 2 is an exponent on x alone, not on the product 9 times x. The power rule only moves an exponent that covers the entire argument.
Trap
The tempting move: you see an exponent anywhere inside and drag it out front.
\[ \log_3\!\left(9x^{2}\right) \;\overset{?}{=}\; 2\log_3(9x) \]
Pull the 2 out in front of everything
Why: The mistake: the 2 is an exponent on x alone, not on the product 9 times x. The power rule only moves an exponent that covers the entire argument.
\[ 2\log_3(9x) \quad \text{\textsf{(false)}} \]
Test both expressions at x equals 3
Why: The original becomes the base-3 log of 9 times 9, which is the log of 81, which is 4. The rewritten version becomes 2 times the log of 27, which is 2 times 3, which is 6. Four does not equal six, so the move was illegal.
\[ \log_3(81) = 4 \qquad \text{but} \qquad 2\log_3 27 = 6 \]
The fix: split the product first with the product rule, then apply the power rule only to the factor that actually carries the exponent.
\[ \log_3\!\left(9x^{2}\right) = \log_3 9 + \log_3 x^{2} \]
Now move the exponent off the single factor
Why: The 2 covers all of x, so it may come out front of that log alone. And the base-3 log of 9 is just 2, because 3 squared is 9.
\[ = 2 + 2\log_3 x \]
Verify at the same test value x equals 3
Why: Substituting gives 2 plus 2 times the base-3 log of 3, which is 2 plus 2 times 1, which is 4. That matches the direct value of the original expression, the log of 81. The corrected expansion agrees.
\[ 2 + 2\log_3 3 = 2 + 2 = 4 = \log_3 81 \ \checkmark \]
Translation
\( 2\log_3(9x) \quad \text{\textsf{(false)}} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Three rules and four identities. Everything else in this deck is these seven lines applied to messier arguments.
| Name | Statement in words |
|---|---|
| Product rule | the log of a product is the sum of the logs |
| Quotient rule | the log of a quotient is the numerator's log minus the denominator's log |
| Power rule | an exponent on the whole argument becomes a multiplier out front |
| Log of 1 | the log of 1 is zero, in every base |
| Log of the base | the log of the base itself is 1 |
| Log of a power of the base | the log of the base raised to a power is that power |
| Base raised to a log | the base raised to a log of a number returns that number |
\[ \log_b(MN) = \log_b M + \log_b N \qquad \log_b\!\left(\tfrac{M}{N}\right) = \log_b M - \log_b N \]
\[ \log_b\!\left(M^{p}\right) = p\log_b M \qquad \log_b 1 = 0 \qquad \log_b b = 1 \]
\[ \log_b\!\left(b^{\,x}\right) = x \qquad b^{\,\log_b x} = x \]
And the one thing that is not on the card, because it is not true: there is no rule for the log of a sum or the log of a difference. If you see a plus sign inside, you are stuck until you factor or combine it.
Section
Part 5
Concept
Every homework problem in this part is one of exactly two requests, and both use the same three rules - just read left to right or right to left.
expand — Rewrite one logarithm of a complicated argument as a sum and difference of simpler logarithms, with no products, quotients, or exponents left inside.
condense — Collapse a sum and difference of logarithms back into a single logarithm of one argument, with no coefficients left in front.
Expanding is what you do to understand an expression. Condensing is what you do to solve an equation, because a single log can be converted to exponential form.
Both require the same base on every log involved. Two logs with different bases cannot be combined at all until change of base fixes that - which is the next part.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of logarithm, argument, common logarithm, vertical asymptote, expand as Logarithmic Functions and Properties of Logs uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Constraint
Discussion prompt
Run Pattern: how to expand a logarithm with this step confiscated:
Distribute the minus into every term that came from the denominator.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Work from the outside in, biggest structure first. Doing it in the other order is where the mistakes come from.
Then check your work by condensing back. If you do not land on the original argument, one of the moves was illegal.
A finished expansion has no parentheses containing products or quotients, and no exponents left inside any log.
Worked example
Expand it completely. Assume all variables are positive.
\[ \log_5\!\left(\frac{x^{3}y}{25}\right) \]
Split the fraction with the quotient rule
Why: The biggest structure is the division bar, so it goes first. The numerator's log comes first, the denominator's is subtracted.
\[ = \log_5\!\left(x^{3}y\right) - \log_5 25 \]
Split the remaining product with the product rule
Why: The numerator is x cubed times y, a product of two factors, so it becomes a sum of two logs.
\[ = \log_5 x^{3} + \log_5 y - \log_5 25 \]
Move the exponent out front with the power rule
Why: The 3 sits on all of x, so it may become a multiplier in front of that log alone. The y has no exponent, so nothing moves there.
\[ = 3\log_5 x + \log_5 y - \log_5 25 \]
Evaluate the numeric log
Why: 25 is 5 squared, so its base-5 log is the plain number 2. Leaving it as a log would be an unfinished expansion.
\[ = 3\log_5 x + \log_5 y - 2 \]
Verify with the test values x equals 5 and y equals 5
Why: The original becomes the base-5 log of 125 times 5 over 25, which is 625 over 25, which is 25 - and the base-5 log of 25 is 2. The expansion becomes 3 times 1 plus 1 minus 2, which is also 2. Both sides give 2, so the expansion is faithful.
\[ \log_5\!\left(\tfrac{125\cdot 5}{25}\right) = \log_5 25 = 2 \qquad 3(1) + 1 - 2 = 2 \ \checkmark \]
Picture it
Animation
Shows: Quotients and powers — a rendered Manim animation.
Rendered with Manim.
Takeaway: The power rule is what makes logs able to solve for an exponent.
Check
Quotient first, then the product in the denominator, then the power. Assume the variables are positive.
\[ \log_2\!\left(\frac{x^{5}}{8y}\right) \]
Check your understanding
Which is the complete, correct expansion?
Answer: A
Why: The quotient rule gives the log of x to the fifth minus the log of 8y. Splitting the denominator gives the log of 8 plus the log of y, and the subtraction applies to both, giving minus 3 minus the log of y. Test with x equals 2 and y equals 2: the original is the base-2 log of 32 over 16, which is 1, and the expansion gives 5 minus 3 minus 1, which is also 1.
Ranking
Put in order
These are the steps of Pattern: how to condense into one logarithm, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Condensing is the expansion run backwards, and the order reverses too: coefficients first, structure second.
Step 1 is the one people skip, and skipping it is the single most common condensing error. A coefficient in front is not allowed to survive.
Then check by expanding your single log back out. You should recover the expression you started from, term for term.
Edge cases
Discussion prompt
Pattern: how to condense into one logarithm works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Condensing is the expansion run backwards, and the order reverses too: coefficients first, structure second.
Worked example
Write it as a single logarithm with coefficient 1.
\[ 3\log x + \tfrac{1}{2}\log y - 2\log z \]
Push every coefficient up as an exponent
Why: The power rule read right to left. Each multiplier climbs back onto its own argument, and nothing is left in front of any log.
\[ = \log x^{3} + \log y^{1/2} - \log z^{2} \]
Rewrite the fractional exponent as a radical
Why: The one-half power is a square root. Doing this now keeps the final argument readable.
\[ = \log x^{3} + \log \sqrt{y} - \log z^{2} \]
Combine the two added logs into one product
Why: The product rule read right to left: a sum of logs is the log of the product of the arguments.
\[ = \log\!\left(x^{3}\sqrt{y}\right) - \log z^{2} \]
Send the subtracted log into the denominator
Why: The quotient rule read right to left. Whatever was being subtracted lands underneath the division bar.
\[ = \log\!\left(\frac{x^{3}\sqrt{y}}{z^{2}}\right) \]
Verify with the test values x equals 10, y equals 100, z equals 10
Why: The original becomes 3 times 1, plus one half times 2, minus 2 times 1, which is 3 plus 1 minus 2, equal to 2. The condensed form becomes the common log of 1000 times 10 over 100, which is the log of 100, which is also 2. The two agree, so the condensing was legal.
\[ 3(1) + \tfrac{1}{2}(2) - 2(1) = 2 \qquad \log\!\left(\tfrac{1000\cdot 10}{100}\right) = \log 100 = 2 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "condensing three common logs", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The original becomes 3 times 1, plus one half times 2, minus 2 times 1, which is 3 plus 1 minus 2, equal to 2. The condensed form becomes the common log of 1000 times 10 over 100, which is the log of 100, which is also 2. The two agree, so the condensing was legal.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: you are condensing, you see a 2 in front, and you slide it inside as a multiplier.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The mistake: the power rule says a coefficient came from an exponent, so it must go back up as an exponent - never in as a factor.
The fix: send the coefficient up as an exponent on its own argument first, before combining anything.
Why: The mistake: the power rule says a coefficient came from an exponent, so it must go back up as an exponent - never in as a factor.
Trap
The tempting move: you are condensing, you see a 2 in front, and you slide it inside as a multiplier.
\[ 2\log_3 x + \log_3 y \;\overset{?}{=}\; \log_3(2xy) \]
Multiply the argument by 2 and combine
Why: The mistake: the power rule says a coefficient came from an exponent, so it must go back up as an exponent - never in as a factor.
\[ \log_3(2xy) \quad \text{\textsf{(false)}} \]
Test both at x equals 3 and y equals 3
Why: The original gives 2 times 1 plus 1, which is 3. The rewritten version gives the base-3 log of 18. But 3 cubed is 27, not 18, so that log is not 3. The two expressions are genuinely different.
\[ 2\log_3 3 + \log_3 3 = 3 \qquad \text{but} \qquad \log_3 18 \ne 3 \]
The fix: send the coefficient up as an exponent on its own argument first, before combining anything.
\[ 2\log_3 x + \log_3 y = \log_3 x^{2} + \log_3 y \]
Now combine the sum into a single product
Why: With no coefficients left in front, the product rule applies cleanly and the arguments simply multiply.
\[ = \log_3\!\left(x^{2}y\right) \]
Verify at the same test values
Why: At x equals 3 and y equals 3 the argument becomes 9 times 3, which is 27, and the base-3 log of 27 is 3 because 3 cubed is 27. That matches the original value of 3 exactly, unlike the illegal version.
\[ \log_3\!\left(3^2 \cdot 3\right) = \log_3 27 = 3 \ \checkmark \]
Check
Three of these four statements are genuine logarithm properties. One is a look-alike that is false. Assume all letters stand for positive numbers.
Check your understanding
Which statement is NOT a valid property of logarithms?
Answer: B
Why: There is no property for the log of a sum. Test it in base 2 with M and N both equal to 8: the left side is the log of 16, which is 4, while the right side is 3 plus 3, which is 6. The other three are the genuine product, quotient, and power rules.
Section
Part 6
Concept
There is a LOG button for base 10 and an LN button for base e. There is no button for base 7. So how do you evaluate a base-7 log?
You rewrite it as a ratio of two logs in a base you do have.
\[ \log_b x = \frac{\log_a x}{\log_a b} \]
The new base a can be anything legal - but in practice you pick 10 or e, because those are the two your calculator can actually compute.
\[ \log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b} \]
Say it as a slogan so the order never slips: the argument goes on top, the base goes on the bottom. The base sinks, because a base always sits low.
Explain it
Discussion prompt
Explain Your calculator only knows two bases to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
You rewrite it as a ratio of two logs in a base you do have.
Intuition
This is not a formula to memorize blindly. It falls out of the definition in three moves, and seeing that makes it impossible to invert by accident.
Name the answer and convert to exponential form
Why: Let y be the log we want. By the definition, the base raised to y produces the argument.
\[ y = \log_b x \quad \Longleftrightarrow \quad b^{\,y} = x \]
Take a log of both sides, in any base you like
Why: An equation stays true if you apply the same function to both sides. Choose a base your calculator knows.
\[ \log_a\!\left(b^{\,y}\right) = \log_a x \ \Longrightarrow \ y\log_a b = \log_a x \]
Divide by the coefficient
Why: The power rule brought y down as a multiplier. Dividing isolates it, and the log of the base is what ends up in the denominator.
\[ y = \frac{\log_a x}{\log_a b} \]
The log of the base landed downstairs because it was the thing multiplying y. That is the memory hook: the base was the coefficient, so the base gets divided out.
Notation
Annotate
From Why the formula is true, in three lines — read this one piece at a time. What is each part doing?
On: \( y = \frac{\log_a x}{\log_a b} \)
Estimation
Predict first
Evaluate it using natural logs. This one comes out exactly, so we can check it by hand.
Commit before you compute: what does Worked example: change of base with an exact answer come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by raising the base to the answer
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. 8 to the five-thirds means the cube root of 8, raised to the fifth.
Worked example
Evaluate it using natural logs. This one comes out exactly, so we can check it by hand.
\[ \log_8 32 \]
Apply the formula with the argument on top
Why: The argument 32 goes in the numerator, the base 8 in the denominator. Reversing these is the classic error.
\[ \log_8 32 = \frac{\ln 32}{\ln 8} \]
Write both arguments as powers of 2
Why: 32 is 2 to the fifth and 8 is 2 cubed. Rewriting exposes structure that a decimal answer would hide.
\[ = \frac{\ln 2^{5}}{\ln 2^{3}} \]
Bring both exponents out front with the power rule
Why: Each exponent covers its whole argument, so each becomes a multiplier.
\[ = \frac{5\ln 2}{3\ln 2} \]
Cancel the common factor
Why: The natural log of 2 is a nonzero number appearing in both numerator and denominator, so it divides out and leaves a clean fraction.
\[ = \frac{5}{3} \]
Verify by raising the base to the answer
Why: 8 to the five-thirds means the cube root of 8, raised to the fifth. The cube root of 8 is 2, and 2 to the fifth is 32 - exactly the argument. The change of base did not distort anything.
\[ 8^{5/3} = \left(\sqrt[3]{8}\right)^{5} = 2^{5} = 32 \ \checkmark \qquad \boxed{\log_8 32 = \tfrac{5}{3}} \]
Picture it
Animation
Shows: Change of base — a rendered Manim animation.
Rendered with Manim.
Takeaway: Which is how a calculator with only ln evaluates any base.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: you remember there is a fraction of two logs, but not which one goes on top.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The mistake: the base was written first in the symbol, so it feels like it should be written first in the fraction.
The fix: argument on top, base on the bottom - always. The base sits low in the symbol, so it sits low in the fraction.
Why: The mistake: the base was written first in the symbol, so it feels like it should be written first in the fraction. Reading order is not fraction order.
Trap
The tempting move: you remember there is a fraction of two logs, but not which one goes on top.
\[ \log_2 8 \;\overset{?}{=}\; \frac{\log 2}{\log 8} \]
Put the base on top and the argument underneath
Why: The mistake: the base was written first in the symbol, so it feels like it should be written first in the fraction. Reading order is not fraction order.
\[ \frac{\log 2}{\log 8} = \frac{\log 2}{3\log 2} = \frac{1}{3} \quad \text{\textsf{(false)}} \]
Test the answer against the definition
Why: If the value were one third, then 2 to the one-third would have to equal 8. But 2 to the one-third is the cube root of 2, a number just over 1 - nowhere near 8. The answer came out as the reciprocal of the truth.
\[ 2^{1/3} = \sqrt[3]{2} \ne 8 \]
The fix: argument on top, base on the bottom - always. The base sits low in the symbol, so it sits low in the fraction.
\[ \log_2 8 = \frac{\log 8}{\log 2} \]
Simplify using the power rule
Why: 8 is 2 cubed, so its log is 3 times the log of 2. That common factor cancels and leaves a whole number.
\[ \frac{\log 8}{\log 2} = \frac{3\log 2}{\log 2} = 3 \]
Verify against the definition and sanity-check the size
Why: 2 cubed is 8, so 3 is correct. And a size check catches this error every time: the argument 8 is bigger than the base 2, so the log must be greater than 1 - one third never had a chance.
\[ 2^{3} = 8 \ \checkmark \qquad 8 > 2 \ \Longrightarrow \ \log_2 8 > 1 \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Prediction
Predict first
What is the value of the base-9 logarithm of 27?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: three halves
Why: Change of base gives the natural log of 27 over the natural log of 9, which is 3 times the log of 3 over 2 times the log of 3, so the value is three halves. Verify: 9 to the three-halves is the square root of 9, cubed, which is 3 cubed, which is 27.
Check
Write both numbers as powers of 3 before you reach for a calculator.
\[ \log_9 27 = \;? \]
Check your understanding
What is the value of the base-9 logarithm of 27?
Answer: A
Why: Change of base gives the natural log of 27 over the natural log of 9, which is 3 times the log of 3 over 2 times the log of 3, so the value is three halves. Verify: 9 to the three-halves is the square root of 9, cubed, which is 3 cubed, which is 27.
Section
Part 7
Concept
The quietest sound you can hear and a jet engine differ by a factor of about a trillion. Plotting both on one linear axis is hopeless.
A log scale fixes that by measuring exponents instead of amounts. A trillion-fold range becomes a range of 12.
\[ \log\left(10^{12}\right) = 12 \]
The payoff, and the thing to remember: on a log scale, equal steps mean equal multiplications, not equal additions.
| Scale | Measures | One whole step means |
|---|---|---|
| decibels | sound intensity | ten decibels is ten times the intensity |
| pH | hydrogen ion concentration | one pH unit is ten times more acidic |
| Richter | earthquake intensity | one magnitude is ten times the intensity |
Comparison
Comparison matrix
From Why the real world uses log scales: refill the One whole step means column from what you know. The rest of the table is as it appeared.
| Scale | Measures | One whole step means |
|---|---|---|
| decibels | sound intensity | ten decibels is ten times the intensity |
| pH | hydrogen ion concentration | one pH unit is ten times more acidic |
| Richter | earthquake intensity | one magnitude is ten times the intensity |
Picture it
Animation
Shows: Why log scales are used — a rendered Manim animation.
Rendered with Manim.
Takeaway: Taking logs turns an unplottable curve into a straight line.
Concept
Loudness is reported as a comparison with the faintest sound a healthy human ear can detect, called the threshold intensity.
\[ I_0 = 10^{-12} \ \text{watts per square meter} \]
The decibel level is ten times the common log of how many times bigger the sound is than that threshold.
\[ D = 10\log\!\left(\frac{I}{I_0}\right) \]
Notice the ratio inside. The units cancel, which is exactly why a log can be taken - you may only take a log of a pure number.
And notice the threshold itself scores zero, because the ratio is 1 and the log of 1 is zero. The scale starts where hearing starts.
Analogy
Discussion prompt
Explain The decibel formula by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Loudness is reported as a comparison with the faintest sound a healthy human ear can detect, called the threshold intensity.
Step zero
Discussion prompt
Worked example: a decibel computation — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the formula and substitute both intensities
Answer:
Worked example
A busy street corner measures this sound intensity. Find its decibel level.
\[ I = 10^{-4} \ \text{watts per square meter} \]
Write the formula and substitute both intensities
Why: Always substitute into the ratio first and simplify it before taking any log. The log is the last operation, not the first.
\[ D = 10\log\!\left(\frac{10^{-4}}{10^{-12}}\right) \]
Simplify the ratio by subtracting exponents
Why: Dividing powers of the same base subtracts the exponents. Negative 4 minus negative 12 is negative 4 plus 12, which is 8.
\[ \frac{10^{-4}}{10^{-12}} = 10^{-4 - (-12)} = 10^{8} \]
Take the common log of a power of 10
Why: This is the round-trip identity: the base-10 log of 10 raised to a power is just that power. No calculator needed.
\[ \log\left(10^{8}\right) = 8 \]
Multiply by 10 and attach the unit
Why: The formula's leading 10 converts bels to decibels. State the answer the way a test would ask for it, with units.
\[ D = 10 \cdot 8 = 80 \ \text{decibels} \]
Verify by running the formula backwards from 80 decibels
Why: If the level is 80, then the log of the ratio is 8, so the ratio is 10 to the eighth. Multiplying by the threshold gives 10 to the eighth times 10 to the negative 12, which is 10 to the negative 4 - exactly the intensity we started with. It also passes a common-sense check: 80 decibels is city-traffic loud.
\[ 10^{8} \cdot 10^{-12} = 10^{-4} \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a decibel computation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If the level is 80, then the log of the ratio is 8, so the ratio is 10 to the eighth. Multiplying by the threshold gives 10 to the eighth times 10 to the negative 12, which is 10 to the negative 4 - exactly the intensity we started with. It also passes a common-sense check: 80 decibels is city-traffic loud.
Concept
Chemists measure acidity by the concentration of hydrogen ions, a number that ranges over many powers of ten. So they report its log - with a minus sign so the answers come out positive.
\[ \text{pH} = -\log\left[H^{+}\right] \]
The minus sign is there because the concentrations are tiny, so their logs are negative. Flipping the sign gives a friendly scale from about 0 to 14.
Because it is a log, a lower pH means a more acidic solution, and each whole unit down is ten times more acidic.
\[ \text{pH } 3 \ \text{is ten times more acidic than pH } 4 \]
The Richter magnitude does the same job for earthquakes. Comparing two quakes is a subtraction of magnitudes and then a power of ten.
\[ \frac{I_1}{I_2} = 10^{\,M_1 - M_2} \]
Counterexample
Discussion prompt
Chemists measure acidity by the concentration of hydrogen ions, a number that ranges over many powers of ten. So they report its log - with a minus sign so the answers come out positive.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The minus sign is there because the concentrations are tiny, so their logs are negative. Flipping the sign gives a friendly scale from about 0 to 14.
Ranking
Put in order
Put the moves of Worked example: the pH of a sample into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. 0.0001 is one ten-thousandth, and ten thousand is 10 to the fourth, so the reciprocal is 10 to the negative 4.
Worked example
A juice sample has this hydrogen ion concentration, in moles per liter. Find its pH and say whether it is acidic.
\[ \left[H^{+}\right] = 0.0001 \]
Rewrite the decimal as a power of 10
Why: 0.0001 is one ten-thousandth, and ten thousand is 10 to the fourth, so the reciprocal is 10 to the negative 4. Decimals hide the exponent that the log wants.
\[ 0.0001 = \frac{1}{10^{4}} = 10^{-4} \]
Substitute into the pH formula
Why: The formula is the negative of the common log of the concentration. Keep the minus sign outside for now so it does not get lost.
\[ \text{pH} = -\log\left(10^{-4}\right) \]
Use the round-trip identity on the inner log
Why: The common log of 10 raised to a power returns that power, so the inner log is negative 4.
\[ \log\left(10^{-4}\right) = -4 \]
Apply the outer minus sign and interpret
Why: The negative of negative 4 is 4. Since 4 is below the neutral value of 7, the sample is acidic - which matches what juice tastes like.
\[ \text{pH} = -(-4) = 4 \]
Verify by running the formula backwards from pH 4
Why: If the pH is 4, then the negative log of the concentration is 4, so the log of the concentration is negative 4, so the concentration is 10 to the negative 4, which is 0.0001. That is exactly the value we were given, so the answer checks.
\[ -\log\left[H^{+}\right] = 4 \ \Longrightarrow \ \left[H^{+}\right] = 10^{-4} = 0.0001 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "the pH of a sample", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If the pH is 4, then the negative log of the concentration is 4, so the log of the concentration is negative 4, so the concentration is 10 to the negative 4, which is 0.0001. That is exactly the value we were given, so the answer checks.
Elimination
Eliminate the wrong options
How many times as intense as the magnitude-5 quake is the magnitude-7 quake?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The magnitudes differ by 2, and the intensity ratio is 10 raised to that difference. 10 squared is 100, so the larger quake is 100 times as intense. Verify by stepping up one magnitude at a time: 5 to 6 multiplies by 10, and 6 to 7 multiplies by 10 again, giving 100.
Check
Two quakes are recorded, one at magnitude 7 and one at magnitude 5. Remember that each whole magnitude is one power of ten of intensity.
\[ \frac{I_1}{I_2} = 10^{\,7 - 5} \]
Check your understanding
How many times as intense as the magnitude-5 quake is the magnitude-7 quake?
Answer: A
Why: The magnitudes differ by 2, and the intensity ratio is 10 raised to that difference. 10 squared is 100, so the larger quake is 100 times as intense. Verify by stepping up one magnitude at a time: 5 to 6 multiplies by 10, and 6 to 7 multiplies by 10 again, giving 100.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The Log Is the Exponent · Common Logs, Natural Logs, and Domain · The Graph: a Mirror Image of the Exponential · The Three Properties of Logarithms · Expanding and Condensing · The Change-of-Base Formula. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
One sentence carried this entire deck: a logarithm is an exponent. Every rule you learned is an exponent rule wearing a different hat.
| If you see... | Do this |
|---|---|
| a log to evaluate | ask: the base to what power gives the argument |
| a variable inside a log | set the argument greater than zero for the domain |
| a product, quotient, or power inside | expand with the matching rule, outside structure first |
| coefficients in front of several logs | send each one up as an exponent, then combine |
| a base your calculator lacks | argument on top, base on the bottom |
| a plus sign inside a single log | stop - there is no rule for the log of a sum |
The four errors we hunted: splitting the log of a sum, dragging out only part of an exponent, ignoring a domain violation, and flipping the change-of-base fraction. Every one of them is caught by the same habit - convert your answer back to exponential form and see whether it is true.
Next up: using all of this to actually solve exponential and logarithmic equations, where checking against the domain stops being optional.
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