This deck closes the exponential and logarithmic thread and opens systems. It solves exponential equations by matching bases and by taking logarithms, isolating the exponential factor first, then condenses and converts logarithmic equations and checks every candidate against the domain. From there it covers doubling time and half-life, and moves into two-variable systems by graphing, substitution, and elimination, inconsistent against dependent systems, three-variable and nonlinear systems, systems of inequalities, and applied mixture, break-even, and investment problems. It targets four killer errors: skipping the domain check on a log equation, taking the log of each term instead of the whole side, multiplying only one term during elimination, and calling a dependent system "no solution".
Subject: College Algebra · 139 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 18
Two big finishes in one deck: how to get a variable out of an exponent (and out of a logarithm), and how to solve two or three equations at once.
Objectives
This deck is the payoff for everything you learned about exponentials and logs, and then it opens the last door of the course: systems.
Along the way we will walk straight into the four mistakes that cost the most points on this material, and defuse each one.
Warm-up
Discussion prompt
Before we open Exponential and Log Equations, and Systems of Equations: without looking back, what was the main idea of Logarithmic Functions and Properties of Logs, 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 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.
Section
Part 1
Concept
An exponential equation has the unknown in the exponent. That single fact decides which tool you reach for.
\[ 2^{\,3x-1} = 32 \qquad \text{versus} \qquad 5^{\,x} = 38 \]
The first is friendly: the number on the right is a whole power of the base, so both sides can be written on the same base. The second is not - no whole power of 5 lands on 38.
exponential equation — An equation in which the unknown appears in an exponent. Solve it by matching bases when you can, and by taking a logarithm of both sides when you cannot.
Counterexample
Discussion prompt
An exponential equation has the unknown in the exponent. That single fact decides which tool you reach for.
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 first is friendly: the number on the right is a whole power of the base, so both sides can be written on the same base. The second is not - no whole power of 5 lands on 38.
Intuition
Think of the base as a currency. Two stacks of the same currency are equal exactly when the counts are equal.
Exponentials are the same story: once both sides are written on one base, the bases carry no more information. Everything left to compare lives in the exponents.
\[ b^{M} = b^{N} \iff M = N \qquad (b > 0,\; b \ne 1) \]
This is called the one-to-one property, and it is only allowed because an exponential function never repeats an output.
Analogy
Discussion prompt
Explain Same base means only the exponents can differ 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:
Think of the base as a currency. Two stacks of the same currency are equal exactly when the counts are equal.
Ranking
Put in order
Put the moves of Worked example: matching the bases 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. Doubling from 2 gives 4, 8, 16, 32 - that is five doublings, so 32 is the fifth power of 2.
Worked example
Solve exactly. Read the right-hand side and ask: is that a power of the base on the left?
\[ 2^{\,3x-1} = 32 \]
Rewrite 32 as a power of 2
Why: Doubling from 2 gives 4, 8, 16, 32 - that is five doublings, so 32 is the fifth power of 2. Now both sides sit on one base.
\[ 2^{\,3x-1} = 2^{5} \]
Set the exponents equal to each other
Why: The one-to-one property: with identical bases, the only way the two sides can agree is for the exponents to agree.
\[ 3x - 1 = 5 \]
Solve the linear equation that is left
Why: Add 1 to both sides to get three x equals six, then divide both sides by 3. The exponential part is completely gone.
\[ x = 2 \]
Verify by substituting into the original equation
Why: Three times two is six, minus one is five, and two to the fifth power is thirty-two. The right side is thirty-two. Both sides agree.
\[ 2^{\,3(2)-1} = 2^{5} = 32 \]
Picture it
Animation
Shows: Matching bases avoids logs entirely — a rendered Manim animation.
Rendered with Manim.
Takeaway: If both sides can be written in one base, just equate the exponents.
Concept
\[ 5^{\,x} = 38 \]
The powers of 5 go 5, 25, 125. Thirty-eight is stranded between the second and the third, so no whole exponent works and no rewrite puts both sides on one base.
So the answer is not a nice integer. It still exists - and a logarithm is exactly the tool built to name it.
The move: take the same logarithm of both whole sides. An equation stays true when you apply one function to both sides.
Explain it
Discussion prompt
Explain When the bases refuse to match 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:
The powers of 5 go 5, 25, 125. Thirty-eight is stranded between the second and the third, so no whole exponent works and no rewrite puts both sides on one base.
Intuition
The unknown is stuck upstairs, riding on the exponent. You cannot subtract it down or divide it down - those moves only reach the ground floor.
The power rule of logarithms is the one tool that reaches upstairs. It takes an exponent and sets it out front as a plain multiplier.
\[ \log_b\!\left(M^{\,p}\right) = p \cdot \log_b M \]
Once the unknown is a multiplier instead of an exponent, the rest is the linear-equation solving you have done all semester.
Concept
Any base works - the equation stays true as long as you apply the same log to both sides. Only convenience decides.
\[ \frac{\ln 38}{\ln 5} = \frac{\log 38}{\log 5} \approx 2.260 \]
That equality is just the change-of-base formula seen from the other direction.
Step zero
Discussion prompt
Worked example: taking the log of both sides — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the natural log of each entire side
Answer:
Worked example
Solve exactly, then give a decimal rounded to three places.
\[ 5^{\,x} = 38 \]
Take the natural log of each entire side
Why: Applying one function to both sides preserves equality. Note it is the log of the whole side, not of the pieces inside it.
\[ \ln\!\left(5^{\,x}\right) = \ln 38 \]
Bring the exponent down with the power rule
Why: The exponent applies to the entire argument, so the whole exponent moves out front as a multiplier. The unknown is no longer upstairs.
\[ x \cdot \ln 5 = \ln 38 \]
Divide both sides by the natural log of 5
Why: The natural log of 5 is just a number, about 1.609. Dividing by it isolates the unknown - this is a one-step linear equation now.
\[ x = \frac{\ln 38}{\ln 5} \]
Evaluate the decimal
Why: The natural log of 38 is about 3.6376 and the natural log of 5 is about 1.6094. Dividing gives about 2.2599.
\[ x = \frac{\ln 38}{\ln 5} \approx 2.260 \]
Verify by substituting into the original equation
Why: Five raised to 2.2599 equals the natural base raised to 2.2599 times 1.6094, which is the natural base raised to 3.6376, and that is 38. Both sides agree.
\[ 5^{\,2.2599} = e^{\,2.2599\,\ln 5} = e^{\,3.6376} = 38 \]
Picture it
Animation
Shows: Which move does this equation want? — a rendered Manim animation.
Rendered with Manim.
Takeaway: Identify where the unknown is hiding and the method follows.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The equation has an added constant sitting next to the exponential.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This is the tempting move - it looks like distributing.
Strip the constant off first, so the exponential stands alone.
Why: This is the tempting move - it looks like distributing. But a logarithm is a function, not a factor, and functions do not distribute across a sum.
Trap
The equation has an added constant sitting next to the exponential.
\[ 2^{\,x} + 3 = 11 \]
Take the log of each term separately
Why: This is the tempting move - it looks like distributing. But a logarithm is a function, not a factor, and functions do not distribute across a sum.
\[ \ln\!\left(2^{\,x}\right) + \ln 3 = \ln 11 \]
Solve that and you get a real number - a wrong one
Why: You would get x times the natural log of 2 equals the natural log of 11 minus the natural log of 3, so x is about 1.874.
\[ x = \frac{\ln 11 - \ln 3}{\ln 2} \approx 1.874 \]
Test it and watch it fail
Why: Two raised to 1.874 is about 3.665, and 3.665 plus 3 is about 6.665 - nowhere near 11. The answer is simply wrong.
\[ 2^{\,1.874} + 3 \approx 6.665 \ne 11 \]
Strip the constant off first, so the exponential stands alone.
\[ 2^{\,x} + 3 = 11 \]
Subtract 3 from both sides
Why: Undo the addition while everything is still ordinary arithmetic. Only then is there a clean exponential to take a log of.
\[ 2^{\,x} = 8 \]
Match the bases - eight is the third power of two
Why: No logarithm is even needed once the exponential is alone; the right side is a whole power of the base.
\[ 2^{\,x} = 2^{3} \;\Rightarrow\; x = 3 \]
Test it and watch it work
Why: Two cubed is eight, and eight plus three is eleven, which is exactly the right side of the original equation.
\[ 2^{3} + 3 = 8 + 3 = 11 \]
Notation
Annotate
From Trap: taking the log of each term — read this one piece at a time. What is each part doing?
On: \( \ln\!\left(2^{\,x}\right) + \ln 3 = \ln 11 \)
Concept
Before any logarithm appears, peel away everything wrapped around the exponential: added constants, multipliers, divisors.
It is the same order of operations you use on a two-step linear equation, run backwards - undo addition before multiplication.
\[ 5e^{\,2x} - 4 = 31 \;\longrightarrow\; 5e^{\,2x} = 35 \;\longrightarrow\; e^{\,2x} = 7 \]
Now, and only now, is the exponential alone on its side and ready for a logarithm.
Picture it
Animation
Shows: Isolate the exponential first — a rendered Manim animation.
Rendered with Manim.
Takeaway: Taking logs before isolating produces an unusable mess.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A number sits in front of the exponential.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It looks like the coefficient belongs to the base.
Treat the coefficient as what it is: a multiplier to divide away.
Why: It looks like the coefficient belongs to the base. It does not - the exponent applies only to the 2, so the 3 is a separate factor.
Trap
A number sits in front of the exponential.
\[ 3 \cdot 2^{\,x} = 48 \]
Multiply the 3 into the base to get a base of 6
Why: It looks like the coefficient belongs to the base. It does not - the exponent applies only to the 2, so the 3 is a separate factor.
\[ 6^{\,x} = 48 \]
That produces a wrong number
Why: The natural log of 48 is about 3.871 and the natural log of 6 is about 1.792, so this path reports about 2.161.
\[ x = \frac{\ln 48}{\ln 6} \approx 2.161 \]
Test it in the original equation
Why: Two raised to 2.161 is about 4.47, and three times that is about 13.4, not 48. The wrong base broke the problem.
\[ 3 \cdot 2^{\,2.161} \approx 13.4 \ne 48 \]
Treat the coefficient as what it is: a multiplier to divide away.
\[ 3 \cdot 2^{\,x} = 48 \]
Divide both sides by 3
Why: The 3 multiplies the whole exponential, so dividing removes it and leaves the exponential alone. The exponent is untouched.
\[ 2^{\,x} = 16 \]
Match bases - sixteen is the fourth power of two
Why: With the exponential isolated, the right side is a clean power of the base, so the exponents can be set equal.
\[ 2^{\,x} = 2^{4} \;\Rightarrow\; x = 4 \]
Test it in the original equation
Why: Two to the fourth is sixteen, and three times sixteen is forty-eight, which matches the right side exactly.
\[ 3 \cdot 2^{4} = 3 \cdot 16 = 48 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The 3 multiplies the whole exponential, so dividing removes it and leaves the exponential alone. The exponent is untouched.
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:
It looks like the coefficient belongs to the base. It does not - the exponent applies only to the 2, so the 3 is a separate factor.
Estimation
Predict first
Solve for the unknown exactly, then to four decimal places.
Commit before you compute: what does Worked example: isolate, then take the natural 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 by substituting into the original equation
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. Twice 0.9730 is 1.9459, which is the natural log of 7, so the natural base raised to it is 7.
Worked example
Solve for the unknown exactly, then to four decimal places.
\[ 5e^{\,2x} - 4 = 31 \]
Add 4 to both sides
Why: Undo the subtraction first. The exponential is a single quantity being multiplied by 5 and then reduced by 4, so the 4 comes off first.
\[ 5e^{\,2x} = 35 \]
Divide both sides by 5
Why: Now the exponential stands completely alone, which is the only state in which a logarithm cleanly applies.
\[ e^{\,2x} = 7 \]
Take the natural log of both sides
Why: The natural log is the exact inverse of the natural base, so the left side collapses to just the exponent - no power rule arithmetic needed.
\[ 2x = \ln 7 \]
Divide both sides by 2
Why: A last one-step linear move. This is the exact answer; keep it in this form unless a decimal is requested.
\[ x = \frac{\ln 7}{2} \approx 0.9730 \]
Verify by substituting into the original equation
Why: Twice 0.9730 is 1.9459, which is the natural log of 7, so the natural base raised to it is 7. Then five times seven is thirty-five, minus four is thirty-one - the original right side.
\[ 5e^{\,2(0.9730)} - 4 = 5(7) - 4 = 31 \]
Picture it
Animation
Shows: Undo a log by exponentiating — a rendered Manim animation.
Rendered with Manim.
Takeaway: Rewrite in exponential form and the log disappears.
Missing information
Discussion prompt
Different bases, and the unknown is upstairs on both sides. Logs handle this without breaking a sweat.
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:
Neither base can be rewritten as a power of the other, so matching bases is off the table. One log applied to both whole sides keeps the equation true.
Worked example
Different bases, and the unknown is upstairs on both sides. Logs handle this without breaking a sweat.
\[ 3^{\,x+1} = 5^{\,2x} \]
Take the natural log of both sides
Why: Neither base can be rewritten as a power of the other, so matching bases is off the table. One log applied to both whole sides keeps the equation true.
\[ \ln\!\left(3^{\,x+1}\right) = \ln\!\left(5^{\,2x}\right) \]
Bring both exponents down
Why: The power rule moves an entire exponent out front - so the whole quantity x plus one comes down, in parentheses. Losing those parentheses is the classic slip here.
\[ (x+1)\ln 3 = 2x \ln 5 \]
Distribute on the left
Why: Now it is an ordinary linear equation whose coefficients happen to be logarithms. Treat those logs as plain numbers.
\[ x\ln 3 + \ln 3 = 2x\ln 5 \]
Collect the unknown terms on one side and factor it out
Why: Subtract the x term on the left from both sides, then factor. Gathering the unknown is the same move you use in any variables-on-both-sides equation.
\[ \ln 3 = x\,(2\ln 5 - \ln 3) \]
Divide by the parenthesized coefficient
Why: That coefficient is just a number, about 2.1203. Dividing gives the exact answer as a ratio of logarithms.
\[ x = \frac{\ln 3}{2\ln 5 - \ln 3} \approx 0.5181 \]
Verify by evaluating both sides of the original equation
Why: The left side is 3 raised to 1.5181, which is about 5.301. The right side is 5 raised to 1.0362, which is also about 5.301. They agree.
\[ 3^{\,1.5181} \approx 5.301 \qquad 5^{\,2(0.5181)} \approx 5.301 \]
Picture it
Animation
Shows: Each line of the worked example "an unknown in both exponents", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The left side is 3 raised to 1.5181, which is about 5.301. The right side is 5 raised to 1.0362, which is also about 5.301. They agree.
Pattern
Five moves, in this order, every single time.
There is no domain trouble here: an exponential is positive for every input, so no exponential equation produces an extraneous solution. That changes in Part 2.
Check
Work it on paper before you choose. Isolate first, then decide whether the bases can match.
\[ 7^{\,x} = 250 \]
Check your understanding
Which expression is the exact solution?
Answer: A
Why: Take the natural log of both sides to get x times ln 7 equals ln 250, then divide both sides by ln 7. That gives 5.5215 divided by 1.9459, which is about 2.837. Checking: 7 raised to 2.837 is about 250.
Section
Part 2
Concept
Before you solve a single logarithmic equation, look at what each logarithm is being handed. That quantity is called the argument, and it must be positive.
\[ \log_b(A) \text{ is defined only when } A > 0 \]
The reason is not a rule someone invented. A log answers the question: what exponent puts the base here? No exponent on a positive base ever produces a negative number or zero.
domain of a logarithmic equation — The set of inputs that make every argument of every logarithm in the ORIGINAL equation strictly positive. Solve the inequalities before you solve the equation, and any candidate outside that set is thrown away.
Definition probe
Sort into buckets
Every line below is part of the definition of exponential equation or of domain of a logarithmic equation — one or the other, never both. Put each where it belongs.
Intuition
Condensing two logs into one is a legal move, but it quietly widens the domain.
\[ \log x + \log (x-3) \;\longrightarrow\; \log\big(x(x-3)\big) \]
On the left, both pieces must be positive, so the unknown has to be bigger than 3. On the right, only the product has to be positive - and a product of two negatives is positive, so numbers below zero sneak in.
Your condensed equation is therefore asking a slightly bigger question than the original. Some of its answers belong to the bigger question only. Those are the extraneous ones.
This is why the domain check is not optional politeness. It is the step that undoes the widening.
Concept
If the equation has exactly one logarithm and it is already by itself, you are one rewrite away from an ordinary equation.
\[ \log_b(A) = c \iff b^{\,c} = A \]
Say it out loud the way it reads: the base raised to the answer gives the argument. The logarithm is the exponent.
That rewrite removes the logarithm entirely, and whatever is left is usually linear or quadratic.
Picture it
Animation
Shows: Take logs to reach an exponent — a rendered Manim animation.
Rendered with Manim.
Takeaway: The power rule is what brings the exponent down.
Fill the middle
Fill in the blanks
From Worked example: a single logarithm — finish the line. Write what belongs on the right of the equals sign before you look.
3^2x - 1 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The argument must be positive, so two x minus one is greater than zero, meaning x is greater than one half.
Worked example
Solve, and state the domain restriction as part of the answer.
\[ \log_3(2x - 1) = 4 \]
State the domain before solving
Why: The argument must be positive, so two x minus one is greater than zero, meaning x is greater than one half. Any candidate at or below one half is dead on arrival.
\[ 2x - 1 > 0 \;\Longrightarrow\; x > \tfrac{1}{2} \]
Convert to exponential form
Why: The base is 3 and the log equals 4, so 3 raised to the fourth power equals the argument. The logarithm is now gone.
\[ 3^{4} = 2x - 1 \]
Evaluate the power
Why: Three to the fourth is eighty-one. Do this arithmetic before solving so the numbers stay small.
\[ 81 = 2x - 1 \]
Solve the linear equation
Why: Add 1 to both sides to get eighty-two, then divide both sides by 2.
\[ x = 41 \]
Check the candidate against the domain
Why: Forty-one is comfortably greater than one half, so the argument is positive and the candidate survives.
\[ 2(41) - 1 = 81 > 0 \]
Verify by substituting into the original equation
Why: The log base 3 of 81 asks what exponent turns 3 into 81, and the answer is 4 - exactly the right side of the original equation.
\[ \log_3\big(2(41)-1\big) = \log_3 81 = 4 \]
Picture it
Animation
Shows: Each line of the worked example "a single logarithm", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Forty-one is comfortably greater than one half, so the argument is positive and the candidate survives.
Concept
You cannot convert to exponential form while two logarithms are still lying around. Squash them into one with the product and quotient rules.
\[ \log_b M + \log_b N = \log_b (MN) \qquad \log_b M - \log_b N = \log_b\!\left(\frac{M}{N}\right) \]
A sum of logs becomes the log of a product; a difference becomes the log of a quotient. Both rules require the same base on every log.
Once there is a single logarithm equal to a number, you are back to the previous slide's one-step rewrite.
Worked example
No base is written, so these are common logs, base ten.
\[ \log x + \log (x - 3) = 1 \]
State the domain first
Why: Both arguments must be positive: x greater than zero AND x minus three greater than zero. The stricter one wins, so the unknown must be greater than 3.
\[ x > 0 \;\text{ and }\; x - 3 > 0 \;\Longrightarrow\; x > 3 \]
Condense the sum into one logarithm
Why: The product rule: a sum of two logs with the same base is the log of the product of the arguments.
\[ \log\big(x(x-3)\big) = 1 \]
Convert to exponential form
Why: The base is ten and the log equals one, so ten to the first power equals the argument.
\[ 10^{1} = x(x-3) \;\Longrightarrow\; x^2 - 3x = 10 \]
Write the quadratic in standard form and factor
Why: Subtract 10 from both sides so one side is zero, then find two numbers multiplying to negative ten and adding to negative three: negative five and positive two.
\[ x^2 - 3x - 10 = 0 \;\Longrightarrow\; (x-5)(x+2) = 0 \]
Read off both candidates
Why: The zero-product property gives two numbers. They are candidates, not yet solutions - the domain has the final word.
\[ x = 5 \quad \text{or} \quad x = -2 \]
Check each candidate against the domain
Why: Five is greater than three, so it is legal. Negative two is not - it would ask for the log of a negative number, which does not exist. Discard it.
\[ x = 5 \;\checkmark \qquad x = -2 \;\text{rejected: } \log(-2) \text{ undefined} \]
Verify the survivor in the original equation
Why: The log of 5 plus the log of 2 is the log of 10 by the product rule, and the common log of ten is one - exactly the right side.
\[ \log 5 + \log 2 = \log 10 = 1 \]
Picture it
Animation
Shows: Each line of the worked example "condense, solve, then discard", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Five is greater than three, so it is legal. Negative two is not - it would ask for the log of a negative number, which does not exist. Discard it.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Same equation, same algebra - but the work stops one step too early.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: All of this is correct work. The quadratic really does factor into these two pieces.
Write the domain on the page before you start, and screen the candidates at the end.
Why: All of this is correct work. The quadratic really does factor into these two pieces.
Trap
Same equation, same algebra - but the work stops one step too early.
\[ \log x + \log (x-3) = 1 \]
Condense, convert, and factor
Why: All of this is correct work. The quadratic really does factor into these two pieces.
\[ (x-5)(x+2) = 0 \]
Report both numbers as the solution set
Why: Habit from quadratic equations: both factors give a zero, so both go on the answer line. That habit is exactly what breaks here.
\[ x = 5, \; x = -2 \quad \text{(both reported)} \]
Substitute negative two and watch it collapse
Why: The very first term of the original equation becomes the log of negative two, which has no value at all. Half of the answer is not a number.
\[ \log(-2) + \log(-5) = 1 \;?\; \text{ undefined} \]
Write the domain on the page before you start, and screen the candidates at the end.
\[ \log x + \log (x-3) = 1 \]
Line one of the work: the domain
Why: Every argument positive. Writing it first means it is impossible to forget it later, when you are tired and staring at a factored quadratic.
\[ x > 3 \]
Screen the candidates against that line
Why: Five passes the domain test, negative two fails it. This filter is a required step of the method, not a bonus.
\[ 5 > 3 \;\checkmark \qquad -2 > 3 \;\text{ false} \]
Report only the survivor, and verify it
Why: The log of five plus the log of two is the log of ten, which equals one. One genuine solution, fully checked.
\[ x = 5 \]
Translation
\( \log x + \log (x-3) = 1 \)
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
If a single logarithm on the left equals a single logarithm on the right, in the same base, then the arguments themselves must be equal.
\[ \log_b M = \log_b N \iff M = N \quad (M > 0,\; N > 0) \]
This is the one-to-one property again, now for logarithms. A log function never returns the same output for two different inputs, so equal outputs force equal inputs.
The parenthetical is the whole danger. Dropping the logs erases the positivity requirement from the page, but not from the problem - so the domain check is still mandatory.
Worked example
A single logarithm on each side, same base.
\[ \log_2\left(x^2 - 6\right) = \log_2 (x) \]
State the domain
Why: Both arguments must be positive: x greater than zero, and x squared minus six greater than zero. Together the unknown must exceed the square root of six, about 2.449.
\[ x > 0 \;\text{ and }\; x^2 > 6 \;\Longrightarrow\; x > \sqrt{6} \approx 2.449 \]
Set the arguments equal
Why: Same base, one log per side, so the one-to-one property applies and both logarithms come off at once.
\[ x^2 - 6 = x \]
Move everything to one side and factor
Why: Subtract x from both sides to reach standard form, then look for two numbers multiplying to negative six and adding to negative one: negative three and positive two.
\[ x^2 - x - 6 = 0 \;\Longrightarrow\; (x-3)(x+2) = 0 \]
List the candidates
Why: Three and negative two. Both satisfy the quadratic; neither is a solution until the domain says so.
\[ x = 3 \quad \text{or} \quad x = -2 \]
Check each against the domain
Why: Three is greater than the square root of six, so it lives. Negative two would ask for the log of a negative number on the right side, so it is extraneous.
\[ 3 > \sqrt{6} \;\checkmark \qquad x = -2 \text{ rejected} \]
Verify the survivor in the original equation
Why: Nine minus six is three, so the left side is the log base two of three, and the right side is the log base two of three. Identical - and both arguments are positive.
\[ \log_2\left(3^2 - 6\right) = \log_2 3 = \log_2 (3) \]
Picture it
Animation
Shows: Each line of the worked example "one-to-one, with a casualty", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Three is greater than the square root of six, so it lives. Negative two would ask for the log of a negative number on the right side, so it is extraneous.
Constraint
Discussion prompt
Run Pattern: solving any logarithmic equation with this step confiscated:
Isolate and condense until there is one logarithm on a side - product rule for sums, quotient rule for differences, power rule for coefficients.
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
The order matters, and the last step is never optional.
Extraneous roots are not rare accidents in this topic - they are the norm. Expect one and you will never be caught.
Edge cases
Discussion prompt
Pattern: solving any logarithmic equation 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:
The order matters, and the last step is never optional.
Check
Find the domain first, then solve. Both matter.
\[ \ln x + \ln (x - 2) = \ln 3 \]
Check your understanding
What is the complete solution set?
Answer: A
Why: Condensing gives x times the quantity x minus 2 equals 3, so x squared minus 2x minus 3 equals zero, which factors into x minus 3 times x plus 1. The domain requires x greater than 2, so negative one is extraneous and 3 is the only solution. Check: ln 3 plus ln 1 equals ln 3 plus 0, which is ln 3.
Section
Part 3
Concept
A continuously growing quantity follows this model, where the starting amount is the number in front and the rate is the constant riding with time.
\[ A = A_0 e^{\,rt} \]
To ask when it doubles, set the ending amount to twice the starting amount. The starting amount then cancels off both sides and never comes back.
\[ 2A_0 = A_0 e^{\,rt} \;\Longrightarrow\; 2 = e^{\,rt} \]
That cancellation is the surprising part: a hundred dollars and a hundred thousand dollars take exactly the same time to double at the same rate.
Picture it
Animation
Shows: Solving a growth model for time — a rendered Manim animation.
Rendered with Manim.
Takeaway: Divide first, then take logs — never the other way round.
Step zero
Discussion prompt
Worked example: how long to double at six percent? — 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 final amount to twice the initial amount
Answer:
Worked example
An account grows continuously at an annual rate of six percent. How long until the balance doubles?
\[ A = A_0 e^{\,0.06t} \]
Set the final amount to twice the initial amount
Why: Doubling is a statement about the ratio, so write it as the ratio the model uses. Nothing else in the problem needs a number yet.
\[ 2A_0 = A_0 e^{\,0.06t} \]
Divide both sides by the starting amount
Why: It is a nonzero common factor, so it cancels. This is why no starting balance was given - it cannot matter.
\[ 2 = e^{\,0.06t} \]
Take the natural log of both sides
Why: The natural log is the exact inverse of the natural base, so the right side collapses to the exponent itself.
\[ \ln 2 = 0.06\,t \]
Divide by the rate to get the symbolic answer
Why: Keep this exact form first - it shows that doubling time is the natural log of two divided by the rate, for every continuous growth problem.
\[ t = \frac{\ln 2}{0.06} \]
Now evaluate it numerically
Why: The natural log of two is about 0.69315. Dividing by 0.06 gives about 11.5525 years, or roughly eleven years and seven months.
\[ t \approx 11.55 \text{ years} \]
Verify by putting the time back into the model
Why: Six hundredths times 11.5525 is 0.69315, which is the natural log of two, so the growth factor is the natural base raised to the natural log of two - which is exactly 2. The balance has doubled.
\[ e^{\,0.06(11.5525)} = e^{\,0.69315} = 2 \]
Picture it
Animation
Shows: Each line of the worked example "how long to double at six percent?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Six hundredths times 11.5525 is 0.69315, which is the natural log of two, so the growth factor is the natural base raised to the natural log of two - which is exactly 2. The balance has doubled.
Concept
Decay problems are usually written with the surviving fraction raised to a power counting how many half-lives have gone by.
\[ A = A_0\left(\frac{1}{2}\right)^{t/h} \]
Here the letter in the denominator of the exponent is the half-life: the time it takes for the exponent to reach one, that is, for one halving to finish.
Once again the starting amount cancels, so these problems are always about the fraction remaining, never the raw quantity.
Sorting
Sort into buckets
These are the pieces of Exponential and Log Equations, and Systems of Equations, out of order. Put each one back under the part of the lesson it belongs to.
Fill the middle
Fill in the blanks
From Worked example: dating a bone with carbon-14 — finish the line. Write what belongs on the right of the equals sign before you look.
\ln 0.6 = \frac{t}{5730}\,\ln\!\left(\frac{1}{2}\right)
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The ending amount is 0.6 times the starting amount.
Worked example
Carbon-14 has a half-life of 5730 years. A bone fragment retains 60 percent of its original carbon-14. How old is it?
\[ A = A_0\left(\frac{1}{2}\right)^{t/5730} \]
Translate sixty percent remaining into an equation
Why: The ending amount is 0.6 times the starting amount. Write that and cancel the starting amount from both sides.
\[ 0.6 = \left(\frac{1}{2}\right)^{t/5730} \]
Take the natural log of both sides
Why: The bases cannot be matched - 0.6 is not a whole power of one half - so the logarithm is the only route to the exponent.
\[ \ln 0.6 = \frac{t}{5730}\,\ln\!\left(\frac{1}{2}\right) \]
Solve for the time symbolically
Why: Divide both sides by the natural log of one half, then multiply both sides by 5730. Both logs are negative, so their ratio is positive - a good sign, since time should be.
\[ t = 5730 \cdot \frac{\ln 0.6}{\ln 0.5} \]
Evaluate
Why: The natural log of 0.6 is about negative 0.51083 and the natural log of 0.5 is about negative 0.69315. Their ratio is about 0.73697, and 5730 times that is about 4222.8.
\[ t \approx 4223 \text{ years} \]
Sanity-check the size of the answer
Why: Sixty percent remaining is less than one full halving, so the age must be less than one half-life of 5730 years. It is, at about 4223.
\[ 0 < 4223 < 5730 \]
Verify by substituting the age into the model
Why: Dividing 4222.8 by 5730 gives 0.73697, and one half raised to that power is the natural base raised to 0.73697 times negative 0.69315, which is the natural base raised to negative 0.51083 - and that is 0.6, exactly sixty percent.
\[ \left(\tfrac{1}{2}\right)^{4222.8/5730} = e^{\,-0.51083} = 0.60 \]
Picture it
Animation
Shows: Each line of the worked example "dating a bone with carbon-14", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Dividing 4222.8 by 5730 gives 0.73697, and one half raised to that power is the natural base raised to 0.73697 times negative 0.69315, which is the natural base raised to negative 0.51083 - and that is 0.6, exactly sixty percent.
Elimination
Eliminate the wrong options
How long does the investment take to double, to the nearest hundredth of a year?
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: Taking the natural log of both sides gives 0.05t equals ln 2, so t is 0.69315 divided by 0.05, which is 13.8629 years. Checking: 0.05 times 13.8629 is 0.69315, and the natural base raised to that is exactly 2.
Check
An investment grows continuously at an annual rate of five percent. Solve for the doubling time before you look at the choices.
\[ 2 = e^{\,0.05t} \]
Check your understanding
How long does the investment take to double, to the nearest hundredth of a year?
Answer: A
Why: Taking the natural log of both sides gives 0.05t equals ln 2, so t is 0.69315 divided by 0.05, which is 13.8629 years. Checking: 0.05 times 13.8629 is 0.69315, and the natural base raised to that is exactly 2.
Section
Part 4
Concept
A system is two or more equations that share the same unknowns. A solution is a single choice of values that makes all of them true at the same time.
\[ \begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases} \]
With two unknowns the answer is an ordered pair, not a single number. Reporting only the first coordinate is reporting half an answer.
solution of a system — An ordered pair (or triple) that satisfies EVERY equation in the system simultaneously. Checking it means substituting into all of the original equations, not just the convenient one.
Picture it
Animation
Shows: Check every log solution — a rendered Manim animation.
Rendered with Manim.
Takeaway: Condensing widened the domain, so a check is mandatory.
Intuition
Each linear equation draws every point that satisfies it: a line. Points on the first line satisfy the first equation; points on the second satisfy the second.
So a point satisfying both must lie on both lines - it is an intersection point. That is the entire geometry of a two-variable linear system.
Two lines can meet once, never, or everywhere. Those three pictures are the only three possible answers: one solution, no solution, or infinitely many.
Matching
Match the pairs
From Where do the two graphs meet? — match each one to what it actually does. The descriptions have been shuffled.
Why: Cross once, Never meet, Lie on top are easy to tell apart while they are sitting next to their descriptions and much harder afterwards, which is what this checks.
Estimation
Predict first
Both equations are already solved for the output, which is exactly the form that graphs fastest.
Commit before you compute: what does Worked example: solving by graphing 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 the pair in BOTH original equations
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. First equation: two times two minus one is three.
Worked example
Figure (svg): Coordinate axes with the solid line y equals 2x minus 1 rising steeply and the dashed line y equals negative x plus 5 falling, crossing at the marked point (2, 3).
Both equations are already solved for the output, which is exactly the form that graphs fastest.
\[ \begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases} \]
Graph the first line from its intercept and slope
Why: Start at the point where the vertical intercept is negative one, then step up two and right one repeatedly. That is the solid line.
Graph the second line the same way
Why: Start at a height of five and step down one for every one step right, since the slope is negative one. That is the dashed line.
Read the intersection point
Why: The two lines cross at a single lattice point, two right and three up. Graphing is only trustworthy when the crossing lands on grid marks like this.
\[ (x, y) = (2, 3) \]
Confirm the reading algebraically
Why: Setting the two expressions for the output equal gives two x minus one equals negative x plus five, so three x equals six and x is two, matching the picture.
\[ 2x - 1 = -x + 5 \;\Longrightarrow\; 3x = 6 \;\Longrightarrow\; x = 2 \]
Verify the pair in BOTH original equations
Why: First equation: two times two minus one is three. Second equation: negative two plus five is three. The same output both times, so the pair truly lies on both lines.
\[ 3 = 2(2) - 1 \quad \text{and} \quad 3 = -(2) + 5 \]
Picture it
Animation
Shows: Each line of the worked example "solving by graphing", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: First equation: two times two minus one is three. Second equation: negative two plus five is three. The same output both times, so the pair truly lies on both lines.
Concept
Graphing is honest but imprecise - a crossing at a third of a unit is unreadable. Substitution is exact.
The idea: if one equation already tells you what one unknown equals, put that expression into the other equation everywhere that unknown appears.
The second equation then holds only one unknown, and one-unknown equations are the thing you already know how to solve.
Reach for substitution when a variable is alone, or has a coefficient of one so that solving for it costs nothing.
Ranking
Put in order
Put the moves of Worked example: solving by substitution 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 first equation says those two things are the same number, so swapping one for the other cannot change what is true.
Worked example
The first equation is already solved for one of the unknowns, so substitution is the fastest route.
\[ \begin{cases} y = 3x - 4 \\ 2x + y = 11 \end{cases} \]
Replace the second equation's y with the expression from the first
Why: The first equation says those two things are the same number, so swapping one for the other cannot change what is true.
\[ 2x + (3x - 4) = 11 \]
Combine like terms
Why: Two x plus three x is five x. Now there is a single unknown, and this is an ordinary two-step equation.
\[ 5x - 4 = 11 \]
Solve for the first coordinate
Why: Add four to both sides to get five x equals fifteen, then divide both sides by five.
\[ x = 3 \]
Back-substitute to get the second coordinate
Why: Use the simplest equation available - the one already solved for the other unknown. Three times three minus four is five.
\[ y = 3(3) - 4 = 5 \;\Longrightarrow\; (3, 5) \]
Verify the pair in BOTH original equations
Why: First: three times three minus four is five, and the pair's second coordinate is five. Second: two times three plus five is eleven, which is the required right side. Both check.
\[ 5 = 3(3)-4 \quad \text{and} \quad 2(3) + 5 = 11 \]
Picture it
Animation
Shows: Each line of the worked example "solving by substitution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: First: three times three minus four is five, and the pair's second coordinate is five. Second: two times three plus five is eleven, which is the required right side. Both check.
Concept
When nothing is solved for a single unknown, solving for one first creates fractions. Elimination avoids that entirely.
Two equal quantities added to two other equal quantities give equal results, so you may add two equations side by side.
Arrange for one variable's coefficients to be opposites before you add, and that variable vanishes in the sum.
\[ \begin{aligned} 6x + 4y &= 32 \\ 5x - 4y &= 12 \end{aligned} \qquad \Longrightarrow \qquad 11x = 44 \]
To manufacture those opposites you may multiply an entire equation by any nonzero number - and entire is the load-bearing word.
Worked example
Neither unknown is isolated, and the second equation already carries a negative four in front of the second unknown.
\[ \begin{cases} 3x + 2y = 16 \\ 5x - 4y = 12 \end{cases} \]
Multiply the entire first equation by 2
Why: Two times two is four, which is the opposite of the negative four below it. Every term gets multiplied - the three x, the two y, and the sixteen.
\[ 6x + 4y = 32 \]
Add the two equations term by term
Why: Four y plus negative four y is zero, so that unknown disappears. Six x plus five x is eleven x, and thirty-two plus twelve is forty-four.
\[ 11x = 44 \]
Solve for the first coordinate
Why: Divide both sides by eleven.
\[ x = 4 \]
Back-substitute into an original equation
Why: Use the first original equation: three times four is twelve, so two y equals four and y is two.
\[ 3(4) + 2y = 16 \;\Longrightarrow\; y = 2 \;\Longrightarrow\; (4, 2) \]
Verify the pair in BOTH original equations
Why: First: twelve plus four is sixteen. Second: five times four is twenty, minus four times two is eight, and twenty minus eight is twelve. Both original right sides are matched.
\[ 3(4)+2(2) = 16 \quad \text{and} \quad 5(4)-4(2) = 12 \]
Picture it
Animation
Shows: Each line of the worked example "solving by elimination", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: First: twelve plus four is sixteen. Second: five times four is twenty, minus four times two is eight, and twenty minus eight is twelve. Both original right sides are matched.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The plan is right: scale the first equation so the second unknown cancels.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The eye is on the two y, so it gets doubled to four y and the rest of the equation is copied down unchanged.
Multiplying an equation means multiplying every term on both sides.
Why: The eye is on the two y, so it gets doubled to four y and the rest of the equation is copied down unchanged. This is the single most common elimination error.
Trap
The plan is right: scale the first equation so the second unknown cancels.
\[ \begin{cases} 3x + 2y = 16 \\ 5x - 4y = 12 \end{cases} \]
Multiply only the term you were staring at
Why: The eye is on the two y, so it gets doubled to four y and the rest of the equation is copied down unchanged. This is the single most common elimination error.
\[ 3x + 4y = 16 \quad \text{(only the middle term doubled)} \]
Add and solve as usual
Why: Three x plus five x is eight x, and sixteen plus twelve is twenty-eight, so this path reports three and a half.
\[ 8x = 28 \;\Longrightarrow\; x = 3.5 \]
Test it in the original system and watch it fail
Why: With x equal to 3.5, the first original equation forces y to be 2.75, but then the second gives 17.5 minus 11, which is 6.5 - not the required 12.
\[ 5(3.5) - 4(2.75) = 6.5 \ne 12 \]
Multiplying an equation means multiplying every term on both sides.
\[ \begin{cases} 3x + 2y = 16 \\ 5x - 4y = 12 \end{cases} \]
Distribute the 2 across the whole first equation
Why: An equation is a statement about two equal quantities; you may double both quantities, which doubles every term inside each.
\[ 2(3x + 2y) = 2(16) \;\Longrightarrow\; 6x + 4y = 32 \]
Add and solve
Why: The y terms cancel, eleven x equals forty-four, so x is four, and back-substitution gives y equal to two.
\[ 11x = 44 \;\Longrightarrow\; (4, 2) \]
Test it in the original system and watch it work
Why: Twelve plus four is sixteen for the first equation, and twenty minus eight is twelve for the second. Every original equation is satisfied.
\[ 3(4)+2(2)=16 \quad 5(4)-4(2)=12 \]
Notation
Annotate
From Trap: multiplying only one term of an equation — read this one piece at a time. What is each part doing?
On: \( 3(4)+2(2)=16 \quad 5(4)-4(2)=12 \)
Concept
Sometimes elimination removes both unknowns at once and leaves a bare numeric statement. That is not a dead end - it is the answer, written in an unfamiliar dialect.
| What is left | What it means | Picture |
|---|---|---|
| A false statement, such as 0 = 4 | No solution (inconsistent) | Parallel lines, never meeting |
| A true statement, such as 0 = 0 | Infinitely many solutions (dependent) | One line drawn twice |
| A normal equation in one unknown | Exactly one solution | Two lines crossing once |
So the question is never whether the variables disappeared. It is whether what remains is true or false.
Comparison
Comparison matrix
From When both variables vanish, read the leftovers: refill the What it means column from what you know. The rest of the table is as it appeared.
| What is left | What it means | Picture |
|---|---|---|
| A false statement, such as 0 = 4 | No solution (inconsistent) | Parallel lines, never meeting |
| A true statement, such as 0 = 0 | Infinitely many solutions (dependent) | One line drawn twice |
| A normal equation in one unknown | Exactly one solution | Two lines crossing once |
Hypothesis
Predict first
Worked example: an inconsistent system 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: Multiply the entire first equation by negative 2
Why: That turns two x into negative four x, the opposite of the four x below, and it doubles-and-negates the other two terms as well.
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
Figure (svg): Two parallel lines of equal slope, y equals 2x minus 3 above y equals 2x minus 5, drawn on coordinate axes and never intersecting.
Solve, or show that no solution exists.
\[ \begin{cases} 2x - y = 3 \\ 4x - 2y = 10 \end{cases} \]
Multiply the entire first equation by negative 2
Why: That turns two x into negative four x, the opposite of the four x below, and it doubles-and-negates the other two terms as well.
\[ -4x + 2y = -6 \]
Add the equations
Why: Negative four x plus four x is zero, and two y plus negative two y is zero. Both unknowns leave together, and only the constants remain.
\[ 0 = 4 \]
Read the leftover statement
Why: Zero is not four. The statement is false for every possible pair, so no pair can satisfy both equations. The system is inconsistent.
\[ \text{No solution} \qquad \varnothing \]
Verify the conclusion by comparing the two lines
Why: Solving each equation for the output gives the same slope of two but different intercepts, negative three and negative five. Equal slopes with different intercepts means parallel and distinct, so they truly never meet.
\[ y = 2x - 3 \quad \text{and} \quad y = 2x - 5 \]
Picture it
Animation
Shows: The one-to-one properties — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both hold because these functions never repeat an output.
Concept
If the second equation is just a multiple of the first, it adds no new information. Every point that satisfies one satisfies the other.
\[ x + 3y = 6 \qquad \text{and} \qquad 2x + 6y = 12 \]
The second equation is literally the first one doubled. Graph them and one line lands exactly on top of the other.
The solution set is then infinite, and you describe it with a rule rather than a point: every pair lying on that shared line.
Step zero
Discussion prompt
Worked example: a dependent system — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Multiply the entire first equation by negative 2
Answer:
Worked example
Solve, and describe the full solution set.
\[ \begin{cases} x + 3y = 6 \\ 2x + 6y = 12 \end{cases} \]
Multiply the entire first equation by negative 2
Why: Aiming to cancel the leading terms. Every term is multiplied: the x, the three y, and the six.
\[ -2x - 6y = -12 \]
Add the equations
Why: Both unknowns cancel and so do the constants. Nothing at all is left except a statement about zero.
\[ 0 = 0 \]
Read the leftover statement
Why: Zero does equal zero. The statement is true for every pair, which says the second equation never restricted anything - the system is dependent.
Describe the infinite solution set with a rule
Why: Solve the first equation for one unknown and let the other run free. Every pair of this form is a solution, and there are infinitely many.
\[ \{ (x, y) \;:\; x = 6 - 3y \} \]
Verify with two sample solutions from that rule
Why: Take y equal to zero, giving the pair six and zero: six plus zero is six, and twelve plus zero is twelve. Take y equal to one, giving three and one: three plus three is six, and six plus six is twelve. Both pairs satisfy both original equations.
\[ (6, 0) \;\checkmark \qquad (3, 1) \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a dependent system", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take y equal to zero, giving the pair six and zero: six plus zero is six, and twelve plus zero is twelve. Take y equal to one, giving three and one: three plus three is six, and six plus six is twelve. Both pairs satisfy both original equations.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Elimination wipes out everything, and the page goes blank.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Correct work: negative six x plus six x is zero, two y plus negative two y is zero, and negative eight plus eight is zero.
Ask one question about the leftover line: is it true or false?
Why: Correct work: negative six x plus six x is zero, two y plus negative two y is zero, and negative eight plus eight is zero.
Trap
Elimination wipes out everything, and the page goes blank.
\[ \begin{cases} 3x - y = 4 \\ 6x - 2y = 8 \end{cases} \]
Multiply the first equation by negative 2 and add
Why: Correct work: negative six x plus six x is zero, two y plus negative two y is zero, and negative eight plus eight is zero.
\[ 0 = 0 \]
Conclude no solution because the variables disappeared
Why: The reasoning is 'nothing is left, so nothing works.' But disappearing variables are not the signal - the truth of the leftover statement is.
\[ \text{No solution} \;\; \text{(reported)} \]
One counterexample destroys that claim
Why: Try the pair two and two. First equation: six minus two is four. Second: twelve minus four is eight. That pair solves both, so 'no solution' is provably false.
\[ (2, 2): \; 3(2)-2 = 4, \;\; 6(2)-2(2) = 8 \]
Ask one question about the leftover line: is it true or false?
\[ \begin{cases} 3x - y = 4 \\ 6x - 2y = 8 \end{cases} \]
Get the same leftover statement
Why: Identical algebra - nothing about the arithmetic changes. Only the interpretation does.
\[ 0 = 0 \]
Zero equals zero is TRUE, so the system is dependent
Why: A statement true for every pair means the second equation placed no new restriction. The two equations describe the same line.
\[ \{ (x, y) \;:\; y = 3x - 4 \} \]
Check the claim with a sample pair
Why: Take x equal to two, so y is two. First equation gives four, second gives eight - both correct. Infinitely many such pairs exist, one for every input.
\[ \text{Infinitely many solutions} \]
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.
Pattern
First choose the tool that costs the fewest fractions.
Then execute, and interpret whatever comes out.
Check
Scale the second equation so one variable cancels, then solve. Write your answer as an ordered pair before choosing.
\[ \begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases} \]
Check your understanding
What is the solution of the system?
Answer: A
Why: Multiplying the whole second equation by 3 gives 3x minus 3y equals 3; adding it to the first gives 5x equals 15, so x is 3, and then 3 minus y equals 1 makes y equal 2. Check: 2 times 3 plus 3 times 2 is 12, and 3 minus 2 is 1.
Section
Part 5
Concept
A three-variable linear system is not a new technique. It is the same elimination, run twice, with bookkeeping.
The answer is an ordered triple, and geometrically it is the single point where three planes meet.
The one rule people break: using the same pair of equations twice. That produces a copy of information you already had, not a second independent equation.
Worked example
Solve for all three unknowns.
\[ \begin{cases} x + y + z = 6 \\ 2x - y + 3z = 9 \\ -x + 2y + z = 6 \end{cases} \]
Target the second unknown, because it already appears with opposite signs
Why: The first equation carries a positive one of it and the second carries a negative one. Adding them straight across removes it with no scaling at all.
Add equation one and equation two
Why: The y terms cancel; x plus two x is three x, z plus three z is four z, and six plus nine is fifteen. Call this new two-variable equation A.
\[ \text{(A)}\quad 3x + 4z = 15 \]
Now use a DIFFERENT pair: equations one and three
Why: Multiply the first equation by negative two so its y term becomes negative two y, the opposite of the positive two y in the third equation.
\[ -2x - 2y - 2z = -12 \]
Add that to equation three
Why: The y terms cancel again; negative two x plus negative x is negative three x, negative two z plus z is negative z, and negative twelve plus six is negative six. Call this equation B.
\[ \text{(B)}\quad -3x - z = -6 \]
Solve the two-variable system formed by A and B
Why: Adding A and B cancels the x terms outright: four z plus negative z is three z, and fifteen plus negative six is nine.
\[ 3z = 9 \;\Longrightarrow\; z = 3 \]
Back-substitute into B to get the first unknown
Why: Negative three x minus three equals negative six, so negative three x is negative three and x is one.
\[ x = 1 \]
Back-substitute both into the simplest original equation
Why: The first equation says the three unknowns add to six, and one plus three is four, so the middle unknown is two.
\[ (x, y, z) = (1, 2, 3) \]
Verify the triple in ALL THREE original equations
Why: First: one plus two plus three is six. Second: two minus two plus nine is nine. Third: negative one plus four plus three is six. Every original right side is matched.
\[ 6 = 6, \qquad 9 = 9, \qquad 6 = 6 \]
Concept
When one equation graphs as a curve, substitution is almost always the tool - solve the simpler equation for one variable and feed it into the curve.
Before you compute anything, sketch and predict how many crossings there should be. It tells you how many answers to expect and catches a lost root instantly.
| System | Possible number of real solutions |
|---|---|
| Line and parabola | 0, 1, or 2 |
| Line and circle | 0 (miss), 1 (tangent), or 2 (secant) |
| Circle and parabola | 0, 1, 2, 3, or 4 |
| Two distinct circles | 0, 1, or 2 |
And every answer is still an ordered pair that must satisfy both original equations - a curve changes the shape of the work, not the meaning of a solution.
Picture it
Animation
Shows: Systems with an exponential — a rendered Manim animation.
Rendered with Manim.
Takeaway: Reduce to a single exponential equation before touching logs.
Worked example
Both equations are already solved for the output, so the two expressions must be equal wherever the graphs meet.
\[ \begin{cases} y = x^2 - 2x - 3 \\ y = x - 3 \end{cases} \]
Set the two expressions equal
Why: At an intersection point the two graphs share the same input and the same output, so their output expressions agree there.
\[ x^2 - 2x - 3 = x - 3 \]
Move everything to one side
Why: Subtract x and add 3 to both sides. The constants cancel completely, which is a gift - the quadratic has no constant term.
\[ x^2 - 3x = 0 \]
Factor out the common factor and use the zero-product property
Why: Do not divide both sides by x - that would delete the solution where x is zero. Factoring keeps both roots alive.
\[ x(x - 3) = 0 \;\Longrightarrow\; x = 0 \;\text{ or }\; x = 3 \]
Find each partner output using the simpler equation
Why: The line is easier arithmetic than the parabola. At an input of zero the output is negative three; at an input of three the output is zero.
\[ (0, -3) \qquad (3, 0) \]
Verify both pairs in BOTH original equations
Why: First pair in the parabola: zero minus zero minus three is negative three; in the line: zero minus three is negative three. Second pair in the parabola: nine minus six minus three is zero; in the line: three minus three is zero. Two genuine intersection points, exactly as a line and a parabola may have.
\[ 0^2-2(0)-3 = -3 \quad 3^2-2(3)-3 = 0 \]
Picture it
Animation
Shows: Each line of the worked example "a line meets a parabola", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: First pair in the parabola: zero minus zero minus three is negative three; in the line: zero minus three is negative three. Second pair in the parabola: nine minus six minus three is zero; in the line: three minus three is zero. Two genuine intersection points, exactly as a line and a parabola may have.
Missing information
Discussion prompt
A circle centered at the origin with radius five, and a line cutting through it.
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:
The line already gives the output in terms of the input, so replace the output everywhere it appears in the circle equation.
Worked example
A circle centered at the origin with radius five, and a line cutting through it.
\[ \begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases} \]
Substitute the line's expression into the circle
Why: The line already gives the output in terms of the input, so replace the output everywhere it appears in the circle equation.
\[ x^2 + (x+1)^2 = 25 \]
Square the binomial correctly
Why: The square of a binomial has three terms - the middle one is twice the product. Squaring term by term here would lose the two x and wreck the problem.
\[ x^2 + x^2 + 2x + 1 = 25 \]
Collect and write in standard form
Why: Two x squared plus two x plus one minus twenty-five gives two x squared plus two x minus twenty-four equal to zero.
\[ 2x^2 + 2x - 24 = 0 \]
Divide the whole equation by 2 and factor
Why: Every term shares a factor of two, and dividing by it keeps the same roots while shrinking the numbers. Then find two numbers multiplying to negative twelve and adding to one: four and negative three.
\[ x^2 + x - 12 = 0 \;\Longrightarrow\; (x+4)(x-3) = 0 \]
Get both inputs and their partner outputs
Why: The inputs are negative four and three. Put each into the line, since it is the easier equation: negative four plus one is negative three, and three plus one is four.
\[ (-4, -3) \qquad (3, 4) \]
Verify both pairs in BOTH original equations
Why: Sixteen plus nine is twenty-five and negative three equals negative four plus one. Nine plus sixteen is twenty-five and four equals three plus one. Both points sit on the circle and on the line, so the line is a secant.
\[ (-4)^2+(-3)^2 = 25 \qquad 3^2+4^2 = 25 \]
Picture it
Animation
Shows: Each line of the worked example "a line meets a circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Sixteen plus nine is twenty-five and negative three equals negative four plus one. Nine plus sixteen is twenty-five and four equals three plus one. Both points sit on the circle and on the line, so the line is a secant.
Prediction
Predict first
What is the complete solution set of the system?
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: (4, 12) and (-1, -3)
Why: Setting the expressions equal gives x squared minus 4 equals 3x, so x squared minus 3x minus 4 equals zero, which factors as x minus 4 times x plus 1. The inputs are 4 and negative 1, and the line gives outputs 12 and negative 3. Check: 16 minus 4 is 12, and 1 minus 4 is negative 3.
Check
Set the two expressions equal and solve. Then find the partner output for each input.
\[ \begin{cases} y = x^2 - 4 \\ y = 3x \end{cases} \]
Check your understanding
What is the complete solution set of the system?
Answer: A
Why: Setting the expressions equal gives x squared minus 4 equals 3x, so x squared minus 3x minus 4 equals zero, which factors as x minus 4 times x plus 1. The inputs are 4 and negative 1, and the line gives outputs 12 and negative 3. Check: 16 minus 4 is 12, and 1 minus 4 is negative 3.
Section
Part 6
Concept
Replace the equals sign of a line with an inequality symbol and the solution stops being a line. It becomes everything on one side of that line.
\[ x + y \le 6 \]
The line itself is the boundary. Draw it solid when the symbol includes equality, and dashed when it does not - dashed means the edge is a fence you may approach but not stand on.
For a system of inequalities, shade each one and keep only the overlap. A point must satisfy every inequality to belong to the solution region.
Explain it
Discussion prompt
Explain One inequality shades half the plane 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:
Replace the equals sign of a line with an inequality symbol and the solution stops being a line. It becomes everything on one side of that line.
Intuition
You never have to reason about which side is correct. Pick any point that is not on the boundary, substitute it, and let the arithmetic answer.
If the statement comes out true, shade the side containing that point. If it comes out false, shade the other side. There is no third possibility.
The origin is the ideal test point because the arithmetic is trivial - use it unless the boundary passes through it.
This works because a boundary line splits the plane into exactly two pieces, and every point in a piece behaves the same way.
Analogy
Discussion prompt
Explain One test point settles which side 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:
You never have to reason about which side is correct. Pick any point that is not on the boundary, substitute it, and let the arithmetic answer.
Picture it
Animation
Shows: Which model fits the data? — a rendered Manim animation.
Rendered with Manim.
Takeaway: Check the successive differences and ratios before choosing.
Fill the middle
Fill in the blanks
From Worked example: graphing a system of inequalities — finish the line. Write what belongs on the right of the equals sign before you look.
0 + 0 = 0 \le 6 \;\; \text{true}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Both symbols include equality, so points on the edges do count as solutions.
Worked example
Figure (svg): Coordinate axes with the boundary lines x plus y equals 6 and y equals x minus 2 crossing at the point (4, 2); the wedge-shaped region below the first line and above the second is shaded, and the origin is marked inside it.
Graph the solution region of this system.
\[ \begin{cases} x + y \le 6 \\ y \ge x - 2 \end{cases} \]
Draw both boundary lines solid
Why: Both symbols include equality, so points on the edges do count as solutions. Solid lines say that out loud.
\[ x + y = 6 \qquad y = x - 2 \]
Test the origin in the first inequality
Why: Zero plus zero is zero, and zero is less than or equal to six - a true statement. So shade the side of that line containing the origin.
\[ 0 + 0 = 0 \le 6 \;\; \text{true} \]
Test the origin in the second inequality
Why: Zero is greater than or equal to zero minus two, that is, zero is at least negative two - also true. Shade that side too.
\[ 0 \ge 0 - 2 \;\; \text{true} \]
Keep only the overlap
Why: The two half-planes overlap in the wedge opening to the left, with its corner where the boundaries meet. Solving the two boundary equations together puts that corner at four and two.
\[ \text{corner at } (4, 2) \]
Verify with one point inside and one point outside
Why: Inside, the origin gives zero at most six and zero at least negative two - both true. Outside, the pair five and five gives ten, which is not at most six, so it is correctly excluded. The shading is right.
\[ (0,0): \text{ both true} \qquad (5,5): \; 10 \not\le 6 \]
Picture it
Animation
Shows: Each line of the worked example "graphing a system of inequalities", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Inside, the origin gives zero at most six and zero at least negative two - both true. Outside, the pair five and five gives ten, which is not at most six, so it is correctly excluded. The shading is right.
Concept
Every applied system follows the same skeleton. Name the unknowns with units, then hunt for two independent facts.
The count equation and the value equation are the two independent facts. If you cannot find a second one, you do not yet have a system.
Counterexample
Discussion prompt
Every applied system follows the same skeleton. Name the unknowns with units, then hunt for two independent facts.
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 count equation and the value equation are the two independent facts. If you cannot find a second one, you do not yet have a system.
Estimation
Predict first
A workshop spends 900 dollars in fixed costs plus 12 dollars of materials per lamp, and sells each lamp for 27 dollars. How many lamps must it sell to break even?
Commit before you compute: what does Worked example: break-even point 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 in BOTH original formulas
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. Cost is twelve times sixty plus nine hundred, which is seven hundred twenty plus nine hundred, or 1620 dollars.
Worked example
A workshop spends 900 dollars in fixed costs plus 12 dollars of materials per lamp, and sells each lamp for 27 dollars. How many lamps must it sell to break even?
Define the unknowns with units
Why: Let the input be the number of lamps and the output be dollars. Two quantities depend on it: total cost and total revenue.
\[ \begin{cases} C = 12x + 900 \\ R = 27x \end{cases} \]
Translate break-even into an equation
Why: Breaking even means revenue exactly covers cost - no profit, no loss. That is the statement that turns two formulas into one system to solve.
\[ 27x = 12x + 900 \]
Collect the variable terms
Why: Subtract twelve x from both sides. Twenty-seven minus twelve is fifteen, and that fifteen is the profit per lamp before fixed costs.
\[ 15x = 900 \]
Solve and interpret
Why: Divide both sides by fifteen. Sixty lamps, each contributing fifteen dollars, exactly covers the nine hundred dollars of fixed cost.
\[ x = 60 \text{ lamps} \]
Verify in BOTH original formulas
Why: Cost is twelve times sixty plus nine hundred, which is seven hundred twenty plus nine hundred, or 1620 dollars. Revenue is twenty-seven times sixty, also 1620 dollars. They match, so sixty lamps is the break-even quantity.
\[ C = 12(60)+900 = 1620 \qquad R = 27(60) = 1620 \]
Picture it
Animation
Shows: Cooling toward room temperature — a rendered Manim animation.
Rendered with Manim.
Takeaway: Exponential decay toward a floor that is not zero.
Step zero
Discussion prompt
Worked example: a mixture problem — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Define the unknowns with units
Answer:
Worked example
A lab needs 12 liters of a 50 percent acid solution. It has a 30 percent solution and a 60 percent solution on the shelf. How much of each should it mix?
Define the unknowns with units
Why: Let the first unknown be liters of the 30 percent solution and the second be liters of the 60 percent solution. Naming them by their strength keeps the two straight.
Write the volume equation
Why: The two poured amounts must total the twelve liters wanted. This is the count equation.
\[ x + y = 12 \]
Write the acid equation
Why: Acid contributed is concentration times volume. Thirty percent of the first plus sixty percent of the second must equal fifty percent of twelve liters, which is six liters of pure acid.
\[ 0.30x + 0.60y = 0.50(12) = 6 \]
Substitute using the volume equation
Why: The first equation gives the second unknown as twelve minus the first, and substituting keeps one unknown.
\[ 0.30x + 0.60(12 - x) = 6 \]
Distribute and collect
Why: Six tenths of twelve is 7.2, and 0.30 minus 0.60 is negative 0.30. So negative 0.30 times the unknown plus 7.2 equals 6.
\[ -0.30x + 7.2 = 6 \;\Longrightarrow\; -0.30x = -1.2 \]
Solve and find the partner
Why: Dividing negative 1.2 by negative 0.30 gives 4 liters of the weaker solution, so the rest, eight liters, is the stronger one.
\[ x = 4 \text{ L}, \qquad y = 8 \text{ L} \]
Verify in BOTH original equations
Why: Volume: four plus eight is twelve liters. Acid: 0.30 times 4 is 1.2 liters, 0.60 times 8 is 4.8 liters, and together that is 6 liters - which is exactly fifty percent of twelve. Both facts check.
\[ 4 + 8 = 12 \qquad 1.2 + 4.8 = 6 = 0.50(12) \]
Picture it
Animation
Shows: Each line of the worked example "a mixture problem", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Volume: four plus eight is twelve liters. Acid: 0.30 times 4 is 1.2 liters, 0.60 times 8 is 4.8 liters, and together that is 6 liters - which is exactly fifty percent of twelve. Both facts check.
Ranking
Put in order
Put the moves of Worked example: splitting an investment 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. Let the first unknown be dollars invested at four percent and the second be dollars at seven percent.
Worked example
Someone invests 10000 dollars, part at 4 percent simple annual interest and the rest at 7 percent. After one year the total interest is 580 dollars. How much went into each account?
Define the unknowns with units
Why: Let the first unknown be dollars invested at four percent and the second be dollars at seven percent. Both are amounts of money, not interest.
Write the money-invested equation
Why: The two pieces must add to the whole ten thousand dollars. That is the count equation.
\[ x + y = 10000 \]
Write the interest equation
Why: Interest is rate times principal. Four percent of the first account plus seven percent of the second must total the reported 580 dollars.
\[ 0.04x + 0.07y = 580 \]
Substitute and distribute
Why: The second account holds ten thousand minus the first. Seven percent of ten thousand is 700, and 0.04 minus 0.07 is negative 0.03.
\[ 0.04x + 0.07(10000 - x) = 580 \;\Longrightarrow\; -0.03x + 700 = 580 \]
Solve for the first amount
Why: Subtract 700 from both sides to get negative 0.03 times the unknown equals negative 120, then divide.
\[ x = 4000 \text{ dollars at } 4\% \]
Find the partner amount
Why: The rest of the ten thousand goes into the higher-rate account: ten thousand minus four thousand.
\[ y = 6000 \text{ dollars at } 7\% \]
Verify in BOTH original equations
Why: Money: four thousand plus six thousand is ten thousand dollars. Interest: four percent of 4000 is 160 dollars, seven percent of 6000 is 420 dollars, and 160 plus 420 is 580 dollars. Both conditions are met.
\[ 4000 + 6000 = 10000 \qquad 160 + 420 = 580 \]
Pattern
The same four questions unlock mixture, money, ticket, and break-even problems alike.
Picture it
Animation
Shows: Growth that runs into a limit — a rendered Manim animation.
Rendered with Manim.
Takeaway: Exponential at first, then flattening as it approaches capacity.
Check
A theater sells adult tickets for 12 dollars and student tickets for 8 dollars. It sold 300 tickets and collected 3080 dollars. Set up both equations before choosing.
Check your understanding
How many adult tickets were sold?
Answer: A
Why: The count equation is a plus s equals 300 and the money equation is 12a plus 8s equals 3080. Substituting s equals 300 minus a gives 4a equals 680, so a is 170 and s is 130. Check: 12 times 170 is 2040, 8 times 130 is 1040, and those total 3080 dollars.
Elimination
Eliminate the wrong options
What does that tell you about the system?
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: A leftover statement that is TRUE for every pair means the second equation added no new restriction, so it is a multiple of the first. The system is dependent and every point on that single shared line is a solution.
Check
While solving a two-variable linear system by elimination, both variables cancel and this is all that is left.
\[ 0 = 0 \]
Check your understanding
What does that tell you about the system?
Answer: A
Why: A leftover statement that is TRUE for every pair means the second equation added no new restriction, so it is a multiple of the first. The system is dependent and every point on that single shared line is a solution.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Getting the Variable Out of the Exponent · Logarithmic Equations and the Domain Check · Doubling Time and Half-Life · Systems: Two Equations at Once · Three Variables and Curved Systems · Shaded Regions and Real Problems. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Two big skills closed out in one deck: pulling a variable out of an exponent or a logarithm, and solving several equations at once.
| If you see | Do this first | Never forget |
|---|---|---|
| An unknown in an exponent | Isolate the exponential, then take a log of the whole side | The log applies to the entire side, not to each term |
| Two or more logarithms | Write the domain, then condense | Screen every candidate against that domain |
| Two linear equations | Pick substitution or elimination | Multiply EVERY term when you scale an equation |
| Both variables cancelling | Look at the leftover statement | True means infinitely many, false means none |
| A word problem | Define unknowns with units, then write a count and a value equation | Answer the question that was asked, in a sentence |
The habit that carries all of it: every answer goes back into the original problem before you call it finished.
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