The definition of a logarithm and the equivalence of logarithmic and exponential form, evaluating logarithms including common and natural ones, the inverse properties and finding inverses of exponential and logarithmic functions, and graphing and translating logarithmic functions.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Evaluate Logarithms and Graph Logarithmic Functions
Objectives
Five outcomes. A logarithm is one thing: the exponent you were looking for.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 498-505 — the lesson these objectives are drawn from
Warm-up
You can evaluate 2 squared and 2 cubed. Chapter 7 has not yet asked the reverse question.
Discussion prompt
Two squared is 4 and 2 cubed is 8. For what value of x does 2 to the x equal 6? Can any method from this course so far find it?
Hint: The answer lies between 2 and 3.
Answer:
Nothing so far can produce it. The answer is somewhere between 2 and 3, and no amount of factoring, root-taking or graphing gives it exactly.
\[ x = \log_2 6 \approx 2.585 \]
So mathematicians named it. A logarithm is the exponent you were looking for, and naming it is what makes the reverse question answerable at all.
Concept
For positive b and y with b not 1, the logarithm of y with base b is the exponent that turns b into y. The statements log base b of y equals x and b to the x equals y say exactly the same thing in two notations.
logarithm of y with base b — The exponent x for which b to the x equals y, written log base b of y. It exists whenever b and y are positive and b is not 1.
\[ \log_b y = x \;\Longleftrightarrow\; b^x = y \]
Because the logarithm undoes the exponential, the two are inverse functions — which is where the graphs, the domains and the simplification rules all come from.
Figure (svg): The logarithmic and exponential forms of the same statement, linked by an if-and-only-if
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 499-499
Section
Section 1
Concept
The equation log base b of y equals x is equivalent to b to the x equals y. Converting between them is a matter of rearranging the three numbers, and every logarithm problem can be attacked in whichever form is easier.
\[ \log_b y = x \;\Longleftrightarrow\; b^x = y \]
Two values are worth recognising immediately: the logarithm of 1 is always 0, since any base to the power zero is 1, and the logarithm of b to base b is always 1.
Figure (svg): The logarithmic and exponential forms of the same statement, linked by an if-and-only-if
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 499-499 — Definition of Logarithm with Base b
Picture it
Example 1, each logarithmic statement beside its exponential twin.
Figure (svg): The logarithmic and exponential forms of the same statement, linked by an if-and-only-if
The base stays the base, the logarithm's value becomes the exponent, and the argument becomes the result. Three numbers, rearranged.
Worked example
Example 1, all four parts.
\[ \text{Rewrite } \log_2 8 = 3, \; \log_4 1 = 0, \; \log_{12} 12 = 1, \; \log_{1/4} 4 = -1. \]
Identify the three roles
Why: The subscript is the base, the value of the logarithm is the exponent, and the argument is the result.
First: 2 to the 3 is 8
Why: The base 2 raised to the exponent 3 gives the argument 8.
\[ 2 ^{3} = 8 \]
Second and third: the special values
Why: Any base to the zero is 1, and any base to the first is itself.
\[ 4 ^{0} = 1; 12 ^{1} = 12 \]
Fourth: a negative exponent
Why: One quarter to the negative 1 is 4, since a negative exponent inverts.
\[ (\frac{1}{4}) ^{-1} = 4 \]
Figure (svg): The solution to Worked example rewrite in exponential form shown as a ladder of expressions, one row per algebraic move
\[ 2^3=8, \; 4^0=1, \; 12^1=12, \; \left(\tfrac{1}{4}\right)^{-1}=4 \]
Verify: check the two special cases in general
Why: The logarithm of 1 to any base is 0, because b to the zero is 1 for every allowed b. The logarithm of b to base b is 1, because b to the first is b. Both are worth memorising rather than recomputing, since they appear constantly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 499-499
Matching
The subscript is the base.
Match the pairs
Why: In every pair the base stays put and the other two numbers swap roles. The last two show that a fractional base and a negative exponent go together whenever the argument exceeds 1.
Worked example
Guided Practice 1 to 4.
\[ \text{Rewrite } \log_3 81 = 4, \; \log_7 7 = 1, \; \log_{14} 1 = 0, \; \log_{1/2} 32 = -5. \]
First: base 3, exponent 4
Why: Three to the fourth is 81.
\[ 3 ^{4} = 81 \]
Second and third: the special values again
Why: Seven to the first is 7; 14 to the zero is 1.
\[ 7 ^{1} = 7; 14 ^{0} = 1 \]
Fourth: a fractional base with a negative exponent
Why: One half to the negative 5 is 2 to the fifth.
\[ (\frac{1}{2}) ^{-5} = 32 \]
Note the pattern
Why: A base below 1 with a negative exponent gives a value above 1.
Figure (svg): The solution to Worked example four more conversions shown as a ladder of expressions, one row per algebraic move
\[ 3^4=81, \; 7^1=7, \; 14^0=1, \; \left(\tfrac{1}{2}\right)^{-5}=32 \]
Verify: check the fourth by inverting
Why: One half to the negative 5 is 2 to the fifth, which is 32. So a logarithm to a base below 1 is negative when its argument is above 1 — the reverse of the usual pattern, and the reason those graphs fall rather than rise.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 499-499
Trap
\[ 2^{-3} = \tfrac{1}{8} \]
Rewrite in logarithmic form with 8 as the base
Why: The number that looks largest is taken as the base.
\[ \log_8 \left(\tfrac{1}{2}\right) = -3 \quad \text{(wrong)} \]
Eight to the negative 3 is one over 512, not one half. The base of the logarithm must be the base of the power.
\[ 2^{-3} = \tfrac{1}{8} \;\Longleftrightarrow\; \log_2\tfrac{1}{8} = -3 \]
Keep the base as the base and the exponent as the value
Why: Only the argument and the exponent trade places between the two forms.
\[ b^x = y \;\Longleftrightarrow\; \log_b y = x \]
Exercise 7 of the lesson asks for exactly this correction. Writing the general form above the specific one, and matching the three roles, prevents it.
Fill the middle
Guided Practice 1.
Fill in the blanks
\log_3 81 = 4 \;\Longleftrightarrow\; 3^4} = 81
Why: The value of the logarithm becomes the exponent, so 3 to the fourth is 81. Checking that arithmetic is the fastest confirmation that the conversion was done the right way round.
Sorting
Two logarithms are known without any work.
Sort into buckets
Sort each logarithm.
Recognising the two special cases on sight saves time on nearly every exercise set in this chapter.
Prediction
Commit before reasoning.
Predict first
The definition excludes b equal to 1 and b negative. Why?
Correct: Because 1 to any power is 1, and a negative base has no consistent real powers.
\[ 1^x = 1 \text{ always} \;\Longrightarrow\; \log_1 y \text{ makes no sense} \]
Why: With base 1 the equation 1 to the x equals y has no solution unless y is 1, and then every x works — so the logarithm would be either undefined or ambiguous. A negative base raised to a fractional exponent often has no real value, as Lesson 6.1 showed. Both exclusions are exactly the ones the exponential functions of Lesson 7.1 already carried.
Section
Section 2
Concept
To evaluate a logarithm, ask what exponent turns the base into the argument. The answer may be negative, fractional or zero, since it is an exponent rather than a count.
common logarithm — A logarithm with base 10, written log without a subscript. A natural logarithm has base e and is written ln.
\[ \log_4 64 = 3, \; \log_5 0.2 = -1, \; \log_{36} 6 = \tfrac{1}{2} \]
Base 10 and base e are common enough to have their own notations: log with no subscript means base 10, and ln means base e. Most calculators have keys for both.
Figure (svg): Four logarithms evaluated by asking what power of the base gives the argument
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 500-500 — Evaluate logarithms
Picture it
Example 2, each written as the question it answers.
Figure (svg): Four logarithms evaluated by asking what power of the base gives the argument
The four answers are 3, negative 1, negative 3 and one half. Every kind of exponent appears, which is exactly why a logarithm can take any real value.
Worked example
Example 2, all four parts.
\[ \text{Evaluate } \log_4 64, \; \log_5 0.2, \; \log_{1/5} 125, \; \log_{36} 6. \]
First: what power of 4 gives 64?
Why: Four cubed is 64.
\[ 3 \]
Second: what power of 5 gives 0.2?
Why: Point two is one fifth, which is 5 to the negative 1.
\[ -1 \]
Third: what power of one fifth gives 125?
Why: One fifth to the negative 3 is 5 cubed, which is 125.
\[ -3 \]
Fourth: what power of 36 gives 6?
Why: The square root of 36 is 6, so the exponent is one half.
\[ \frac{1}{2} \]
Figure (svg): The solution to Worked example evaluate four logarithms shown as a ladder of expressions, one row per algebraic move
\[ 3, \; -1, \; -3, \; \tfrac{1}{2} \]
Verify: convert each answer back
Why: Four cubed is 64; 5 to the negative 1 is 0.2; one fifth to the negative 3 is 125; 36 to the one half is 6. All four check. Converting back to exponential form is the natural verification for every logarithm evaluation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 500-500
Sorting
Compare the argument with 1, for a base above 1.
Sort into buckets
Sort each logarithm by the sign of its value.
For a base above 1, the sign of a logarithm is decided entirely by whether its argument is above or below 1 — a useful check before any calculation.
Worked example
Examples 3 and 4, plus Guided Practice 5 to 9.
\[ \text{Evaluate } \log 8, \; \ln 0.3, \; \log_2 32, \; \log_{27} 3, \text{ and find the tornado speed for } d = 220. \]
The two exact ones
Why: Two to the fifth is 32, and 27 to the one third is 3.
\[ 5\text{ and } \frac{1}{3} \]
The two calculator ones
Why: Log 8 is about 0.903 and ln 0.3 is about negative 1.204.
\[ 0.903, -1.204 \]
Check them
Why: Ten to the 0.903 is about 8, and e to the negative 1.204 is about 0.3.
The model
Why: Ninety-three times log 220, plus 65, with log 220 about 2.342.
\[ \text{about } 283 \text{mph} \]
Figure (svg): The solution to Worked example common and natural logarithms shown as a ladder of expressions, one row per algebraic move
\[ 5, \; \tfrac{1}{3}, \; 0.903, \; -1.204; \quad 283 \text{ mph} \]
Verify: sanity-check the two calculator values
Why: Eight is between 1 and 10, so log 8 must be between 0 and 1 — and 0.903 is. Point three is below 1, so its natural logarithm must be negative — and it is. Bracketing a logarithm between two easy values before reading a display catches a mistyped base or argument.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 500-500
Error analysis
A student evaluates a logarithm.
Annotate
On: \( \log_2(-8) = -3, \text{ since } 2^{-3} = -8 \)
The argument of a logarithm must be positive. That single restriction is what gives every logarithmic graph a domain of x greater than zero.
Fill the middle
Example 2d.
Fill in the blanks
\log_1/2 6: \; 36 \text___ 6? \; 36^___} = 6
Why: The square root of 36 is 6, and a square root is the one-half power, so the logarithm is one half. Fractional answers are entirely ordinary here, since the logarithm is just an exponent and exponents may be fractions.
Matching
Two bases have their own symbols.
Match the pairs
Why: The last two rows are the same function written two ways: ln is simply the usual abbreviation for log base e. A missing subscript always means base 10, never base e, which is the convention most often confused.
Prediction
Commit before reasoning.
Predict first
In the tornado model, the path length rises from 150 to 220 miles. What happens to the predicted speed?
Correct: It rises from about 267 to about 283, a change of about 6 percent.
\[ \log 150 \approx 2.176, \quad \log 220 \approx 2.342 \]
Why: A 47 percent longer path buys only a 6 percent higher speed, because a logarithm grows more and more slowly. To add another 16 miles per hour the path would have to grow by another 47 percent, to about 323 miles. That flattening is the defining feature of logarithmic growth and is why logarithms are used for quantities spanning huge ranges.
Section
Section 3
Concept
Because the logarithm with base b is the inverse of the exponential with base b, composing them either way returns the input. That gives two simplification rules and a method for finding inverses.
\[ b^{\log_b x} = x \quad \text{and} \quad \log_b b^x = x \]
These are Lesson 6.4's inverse conditions written for this pair. The definition of a logarithm is precisely the statement that the two functions undo each other.
Figure (svg): Two columns comparing an exponential function with its logarithmic inverse
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 501-501 — Use inverse properties
Picture it
The exponential function beside its logarithmic inverse.
Figure (svg): Two columns comparing an exponential function with its logarithmic inverse
Domain becomes range, range becomes domain, a horizontal asymptote becomes a vertical one, and the point where x is 0 and y is 1 becomes the point where x is 1 and y is 0.
Worked example
Example 5, both parts.
\[ \text{Simplify } 10^{\log 4} \text{ and } \log_5 25^x. \]
First: recognise the pattern
Why: A base raised to a logarithm with the same base returns the argument.
\[ 10 ^{\log 4} = 4 \]
Second: rewrite the argument as a power of the base
Why: Twenty-five is 5 squared.
\[ \log _{5}((5 ^{2}) ^{x}) \]
Second: apply the power of a power rule
Why: Two times x is 2x.
\[ \log _{5}(5 ^{2 x}) \]
Second: apply the other inverse property
Why: A logarithm of the base raised to a power returns that power.
\[ 2 x \]
Figure (svg): The solution to Worked example simplify with the inverse properties shown as a ladder of expressions, one row per algebraic move
\[ 10^{\log 4} = 4; \qquad \log_5 25^x = 2x \]
Verify: check the second at a value
Why: At x equal to 3, 25 cubed is 15,625, and 5 to the sixth is also 15,625 — so the logarithm base 5 is 6, which is 2 times 3. The rewriting step is what made the inverse property applicable, and without it the expression looks unrelated to base 5.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 501-501
Fill the middle
Guided Practice 13.
Fill in the blanks
e^20 = ___
Why: The natural logarithm has base e, so raising e to it returns the argument. That is the property b to the log base b of x equals x with b equal to e, and it works because the two functions are inverses.
Worked example
Example 6 and Guided Practice 14 and 15.
\[ \text{Find the inverses of } y = 6^x, \; y = \ln(x+3), \; y = 4^x, \; y = \ln(x-5). \]
First: use the definition directly
Why: The inverse of an exponential with base 6 is the logarithm with base 6.
\[ y = \log _{6}(x) \]
Second: switch x and y
Why: The equation becomes x equals the natural logarithm of y plus 3.
\[ x = \ln(y + 3) \]
Second: convert to exponential form and solve
Why: E to the x equals y plus 3, so y is e to the x minus 3.
\[ y = e ^{x} - 3 \]
Third and fourth: the same two methods
Why: The inverse of 4 to the x is log base 4; for the last, e to the x plus 5.
\[ \log _{4}(x); e ^{x} + 5 \]
Figure (svg): The solution to Worked example find two inverses shown as a ladder of expressions, one row per algebraic move
\[ \log_6 x, \; e^x-3, \; \log_4 x, \; e^x+5 \]
Verify: check the second by composing
Why: Substituting e to the x minus 3 into the natural logarithm of x plus 3 gives the natural logarithm of e to the x, which is x. Both compositions return x, which is Lesson 6.4's test for a genuine pair of inverses.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 501-501
Trap
\[ 2^{\log_5 x} \]
Cancel the exponential against the logarithm
Why: The pattern base-to-a-logarithm is spotted and applied.
\[ = x \quad \text{(wrong)} \]
At x equal to 25 the logarithm base 5 is 2, so the expression is 2 squared, which is 4 — not 25.
\[ b^{\log_b x} = x \text{ requires the SAME base twice} \]
Check that the two bases match before cancelling
Why: The properties are statements about a function and its own inverse.
\[ 5^{\log_5 x} = x \quad \checkmark, \qquad 2^{\log_5 x} \neq x \]
Lesson 7.5's change of base formula will handle mismatched bases. Until then, matching bases is a precondition rather than a detail.
Matching
Match the bases first.
Match the pairs
Why: The third needed a rewriting step first: 64 is 2 to the sixth, so 64 to the x is 2 to the 6x and the logarithm base 2 returns 6x. Whenever the bases do not match on sight, look for a way to rewrite one as a power of the other.
Ranking
Finding the inverse of a logarithmic function.
Put in order
Why: Step three is the one that is new: once the variables are switched, a logarithm is trapped and converting to exponential form is what frees it. From there the algebra is ordinary. The check is Lesson 6.4's, and it requires both directions.
Prediction
Commit before reasoning.
Predict first
Why does log base b of b to the x equal x, for every x?
Correct: Because the logarithm asks what exponent gives b to the x, and the answer is written there.
\[ \log_b b^x = x \text{ is the definition, restated} \]
Why: The logarithm's whole job is to report the exponent, and in b to the x the exponent is x, plainly visible. So the property is not a rule on top of the definition; it is the definition read aloud. The other property is the same statement from the other side, and it works for every real x — positive, negative or zero.
Section
Section 4
Concept
Because the logarithm is the inverse of the exponential, its graph is the reflection of the exponential's in the line y equals x. The domain is the positive numbers, the range is all real numbers, and the vertical axis is an asymptote.
\[ f(x) = \log_b x, \quad x > 0 \]
The graph rises for a base above 1 and falls for a base below 1, and in both cases it passes through the point where x is 1 and y is 0.
Figure (svg): An exponential graph and its logarithmic inverse, reflections of each other in the line y equals x
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 502-502 — Parent Graphs for Logarithmic Functions
Picture it
Three to the x and log base 3, reflected in the dashed line.
Figure (svg): An exponential graph and its logarithmic inverse, reflections of each other in the line y equals x
The point where x is 0 and y is 1 on one curve becomes the point where x is 1 and y is 0 on the other. The horizontal asymptote becomes a vertical one.
Worked example
Example 7, both parts.
\[ \text{Graph } y = \log_3 x \text{ and } y = \log_{1/2} x. \]
First: choose convenient points
Why: Powers of the base give whole-number outputs: 1, 3 and 9 give 0, 1 and 2.
\[ (1, 0), (3, 1), (9, 2) \]
First: draw the curve
Why: It starts just right of the vertical axis and rises through the points.
Second: choose points for base one half
Why: Powers of one half give 1, 2, 4 and 8 mapping to 0, negative 1, negative 2 and negative 3.
\[ (1, 0), (2, -1), (4, -2), (8, -3) \]
Second: draw the curve
Why: It starts just right of the vertical axis and falls through the points.
Figure (svg): The solution to Worked example graph two logarithmic functions shown as a ladder of expressions, one row per algebraic move
\[ \log_3 x \text{ rises}; \quad \log_{1/2} x \text{ falls} \]
Verify: check the choice of points
Why: Using powers of the base makes every output a whole number, which is why 1, 3 and 9 were chosen for base 3 rather than 1, 2 and 3. For base one half the powers run 1, 2, 4, 8 with outputs 0, negative 1, negative 2, negative 3 — the same trick, giving evenly spaced heights.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 502-502
Sorting
Compare the base with 1.
Sort into buckets
Sort each logarithmic function.
The base decides direction here exactly as it did for the exponential functions of Lessons 7.1 and 7.2 — which is unsurprising, since these are those functions reflected.
Worked example
Guided Practice 16, plus the general statement.
\[ \text{State the domain, range and asymptote of } y = \log_5 x \text{ and of } y = \log_b x \text{ in general.} \]
Find the domain
Why: The argument of a logarithm must be positive.
\[ x > 0 \]
Find the range
Why: Every real number is the logarithm of something, since the exponential takes every positive value.
Find the asymptote
Why: As x approaches zero from the right the outputs run to negative infinity.
Compare with the exponential
Why: Each of these is the exponential's corresponding feature with the roles swapped.
Figure (svg): The solution to Worked example state domains and ranges shown as a ladder of expressions, one row per algebraic move
\[ x > 0, \; \text{range all reals}, \; \text{asymptote } x = 0 \]
Verify: check the asymptote numerically
Why: Log base 5 of one twenty-fifth is negative 2, of one over 3125 is negative 5, and of one over 5 to the hundredth is negative 100. The outputs fall without bound as the input approaches zero, which is what a vertical asymptote means. The input never reaches zero, and neither does the graph.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 502-502
Error analysis
A student sketches a logarithmic function.
Annotate
On: \( y = \log_3 x \text{, drawn for both positive and negative } x \)
The domain restriction is not a convention; it is what the definition requires. It is also the exponential function's range, reflected.
Fill the middle
Example 7a.
Fill in the blanks
y = \log_3 x: \; \text2 x = 9, \; y = ___ \text___ 3^2 = 9
Why: Nine is 3 squared, so the logarithm is 2. Choosing arguments that are powers of the base gives whole-number outputs, which is why 1, 3 and 9 are the natural points to plot rather than 1, 2 and 3.
Comparison
Fill the blanks. Every feature swaps.
Comparison matrix
| Feature | y = b^x | y = log_b(x) |
|---|---|---|
| Domain | all real numbers | x > 0 |
| Range | y > 0 | all real numbers |
| Asymptote | horizontal, the x-axis | vertical, the y-axis |
| Key point | (0, 1) | (1, 0) |
Every row is the same fact with the coordinates reversed, which is exactly what reflecting in the line y equals x does.
Prediction
Commit before reasoning.
Predict first
The exponential function has a horizontal asymptote. Why is the logarithm's vertical?
Correct: Because reflecting in y equals x turns horizontal lines into vertical ones.
\[ y = 0 \;\to\; x = 0 \quad \text{under reflection in } y = x \]
Why: The exponential graph flattens toward the horizontal axis, and reflecting the whole picture in the line y equals x carries that axis onto the vertical one. This is the same reflection argument that turned the horizontal line test into the vertical line test in Lesson 6.4, and it explains the asymptote without any new computation.
Section
Section 5
Concept
To graph the logarithm of x minus h, plus k, translate the parent graph h units horizontally and k units vertically. Because the asymptote is vertical, it moves to the line x equals h, and the domain becomes x greater than h.
\[ y = \log_b(x-h)+k \;\Longrightarrow\; \text{asymptote } x = h \]
This is the reverse of the exponential case, where the asymptote was horizontal and moved with k. Reading which asymptote a family has is what decides which letter moves it.
Figure (svg): A logarithmic graph translated left and up, with its vertical asymptote moved
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 503-503 — Translate a logarithmic graph
Picture it
Example 8: the logarithm base 2 of x plus 3, plus 1.
Figure (svg): A logarithmic graph translated left and up, with its vertical asymptote moved
The three plotted points moved left 3 and up 1, and the asymptote moved from the vertical axis to the line three units to its left. The range is still every real number.
Worked example
Example 8, in the book's two steps.
\[ \text{Graph } y = \log_2(x+3)+1 \text{ and state the domain and range.} \]
Sketch the parent
Why: Log base 2 passes through the points where x is 1, 2 and 4 with outputs 0, 1 and 2.
Read h and k
Why: X plus 3 gives h equal to negative 3, and the constant outside is 1.
\[ h = -3, k = 1 \]
Shift the graph
Why: Left 3 and up 1, carrying the three points.
\[ (-2, 1), (-1, 2), (1, 3) \]
State the asymptote, domain and range
Why: The asymptote moves to the line x equals negative 3.
\[ x > -3;\text{ all reals} \]
Figure (svg): The solution to Worked example translate a logarithmic graph shown as a ladder of expressions, one row per algebraic move
\[ x > -3, \quad \text{asymptote } x = -3 \]
Verify: check one shifted point
Why: At x equal to negative 2 the argument is 1, whose logarithm is 0, so the value is 0 plus 1, which is 1 — matching the shifted point. And the domain condition comes from the argument: x plus 3 must be positive, so x exceeds negative 3.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 503-503
Matching
The asymptote is the line x equals h.
Match the pairs
Why: The constant added outside plays no part at all: the first and third both have one and neither affects the asymptote. Setting the argument equal to zero locates it in one step.
Worked example
Guided Practice 17 and 18.
\[ \text{Graph } y = \log_{1/3}(x-3) \text{ and } f(x) = \log_4(x+1)-2. \]
First: read h and k
Why: X minus 3 gives h equal to 3; there is no constant outside.
\[ h = 3, k = 0 \]
First: state the domain and direction
Why: The argument must be positive, so x exceeds 3; the base is below 1, so the curve falls.
\[ x > 3,\text{ falling} \]
Second: read h and k
Why: X plus 1 gives h equal to negative 1, and the constant is negative 2.
\[ h = -1, k = -2 \]
Second: state the domain and direction
Why: The argument must be positive, so x exceeds negative 1; the base 4 is above 1, so it rises.
\[ x > -1,\text{ rising} \]
Figure (svg): The solution to Worked example two more translations shown as a ladder of expressions, one row per algebraic move
\[ x > 3; \qquad x > -1 \]
Verify: derive each domain from the argument
Why: For the first, x minus 3 must be positive, giving x greater than 3; for the second, x plus 1 must be positive, giving x greater than negative 1. Both agree with the value of h, as they must — the domain of a translated logarithm always begins exactly at h.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 503-503
Trap
\[ y = \log_2(x+3)+1 \]
Put the asymptote at y equals 1
Why: The rule from Lessons 7.1 and 7.2 is carried across.
\[ \text{asymptote } y = 1 \quad \text{(wrong)} \]
The graph passes straight through the height 1 at x equal to negative 2, so that line cannot be an asymptote.
\[ \text{asymptote } x = h = -3 \]
Move the vertical asymptote with h
Why: A logarithmic graph's asymptote is vertical, so a horizontal shift is what moves it.
\[ \text{argument } x+3 > 0 \;\Longrightarrow\; x > -3 \]
Which letter moves the asymptote depends on which way the asymptote runs. Exponential graphs have horizontal asymptotes moved by k; logarithmic graphs have vertical ones moved by h.
Fill the middle
Example 8.
Fill in the blanks
y = \log_2(x+3)+1: \; x+3 > 0 \;\Longrightarrow\; x > -3
Why: The argument must be positive, giving x greater than negative 3 — the same value as h. The domain of a translated logarithm always starts at h and runs right, which mirrors how a square root's domain started at h in Lesson 6.5.
Sorting
It depends on which way the asymptote runs.
Sort into buckets
Sort each family by what moves its asymptote.
Deciding which letter matters is a question about the shape of the parent, not about the algebra of the shift.
Prediction
Commit before reasoning.
Predict first
The parent logarithm has range all real numbers. What is the range after translating?
Correct: Still all real numbers.
\[ \text{range of } \log_b(x-h)+k \text{ is all reals, for every } h \text{ and } k \]
Why: The parent already takes every real value, so shifting it vertically by k simply relabels which input produces which output — every value is still attained. This is the opposite of the exponential case, where the range was one-sided and k moved its boundary. What a translation restricts here is the domain, not the range.
Comparison
Fill the blanks. Inverse functions swap everything.
Comparison matrix
| Feature | Exponential | Logarithmic |
|---|---|---|
| Domain | all reals | x > 0 |
| Asymptote | horizontal, moved by k | vertical, moved by h |
| Growth for a base above 1 | faster and faster | slower and slower |
| Key point | (0, 1) | (1, 0) |
The third row is the practical consequence: exponentials outrun everything and logarithms flatten toward nothing, which is why the pair is so useful for quantities spanning huge ranges.
Pattern
One routine for evaluating, one for graphing.
The argument of a logarithm must always be positive. That single condition supplies every domain in the lesson.
OpenStax Algebra and Trigonometry 2e, §6.3 Logarithmic Functions §6.3
Check
Evaluating. Ask the question.
Check your understanding
Evaluate log_(1/5)(125).
Answer: A
Why: One fifth to the -3 is 5^3, which is 125.
Check
Inverse properties. Bases must match.
Check your understanding
Simplify log_5(25^x).
Answer: A
Why: 25 is 5^2, so 25^x is 5^(2x), and log base 5 of that is 2x.
Check
Translating. Which letter moves the asymptote?
Check your understanding
What are the asymptote and domain of y = log_2(x + 3) + 1?
Answer: A
Why: The argument x + 3 must be positive, so x > -3, and the vertical asymptote sits at that boundary.
Real world
An earthquake's magnitude on the Richter scale is M equals the common logarithm of I over I nought, where I is the quake's intensity and I nought a reference intensity.
Discussion prompt
Find the magnitude of a quake 10,000 times the reference intensity, then find how many times more intense a magnitude 7 quake is than a magnitude 5 one, and explain why the scale is logarithmic.
Hint: Convert to exponential form for the second question.
Answer:
\[ M = \log(10{,}000) = \log(10^4) = 4 \]
\[ M = 7 \;\Longrightarrow\; \tfrac{I}{I_0} = 10^7; \quad M = 5 \;\Longrightarrow\; \tfrac{I}{I_0} = 10^5 \]
A quake 10,000 times the reference has magnitude 4, and a magnitude 7 quake is 100 times more intense than a magnitude 5 one — not 40 percent more.
The scale is logarithmic because earthquake intensities span an enormous range, from barely detectable to catastrophic, over many powers of ten. A logarithm compresses that range into single-digit numbers, so a scale from 1 to 9 covers a factor of a hundred million. That compression is exactly the slow growth seen in the tornado model, put to deliberate use.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does log base 2 of negative 8 have a real value?
Correct: No — no real power of 2 is negative.
\[ 2^x > 0 \text{ for every real } x \;\Longrightarrow\; \log_2(-8) \text{ undefined} \]
Why: Two raised to any real exponent is positive: 2 to the 3 is 8, 2 to the negative 3 is one eighth, and 2 to the zero is 1. Nothing produces negative 8. The definition therefore requires the argument to be positive, and that restriction is exactly what gives every logarithmic graph a domain of x greater than zero and a vertical asymptote at the origin. A negative VALUE of a logarithm is perfectly ordinary; a negative argument is not.
Explain it
They can compute 2 to the fifth and have just seen log base 2 of 32.
Discussion prompt
In four sentences or fewer, explain what a logarithm is, without using the word inverse.
Hint: It answers a question about an exponent.
Answer:
A logarithm is the exponent you need. Writing log base 2 of 32 is asking the question: 2 raised to what power gives 32? The answer is 5, so log base 2 of 32 is 5.
That is all there is to the definition, and every property comes from it. The answer can be a fraction or negative, because exponents can be — log base 36 of 6 is one half, and log base 5 of 0.2 is negative 1.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For converting, write the general form above the specific one and match the three roles. For fractional bases, remember that a base below 1 needs a negative exponent to exceed 1. For cancelling, check the two bases match before doing anything. For asymptotes, set the argument equal to zero. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference for logarithms. Top left: write the definition with the if-and-only-if, then four conversions of your own with the base circled in both forms. Top right: evaluate six logarithms by writing each as the question it asks, including one with a fractional base, one with a negative answer and one with a fractional answer. Middle: sketch y equals 3 to the x and y equals log base 3 of x on one set of axes with the line y equals x dashed, marking the two key points and both asymptotes. Bottom left: write both inverse properties and use each on an expression you invent, including one where the argument has to be rewritten as a power of the base first. Bottom right: graph y equals log base 2 of x plus 3, plus 1, marking three shifted points and the vertical asymptote, and state the domain and range with a sentence explaining where each came from. In a margin, write which letter moves the asymptote for exponentials and which for logarithms.
If your bottom-right domain is not x greater than negative 3, recheck the argument: it is x plus 3 that must be positive, not x.
Recap
Five things, and all of them come from one definition.
| If you see | Then |
|---|---|
| log_b(y) = x | It means b^x = y |
| An argument of 1 | The logarithm is 0 |
| An argument equal to the base | The logarithm is 1 |
| log with no subscript | Base 10 |
| ln | Base e |
| b raised to a log with the same base | The two cancel |
| log_b(x - h) + k | Asymptote x = h, domain x > h |
Lesson 7.5 develops the properties of logarithms — product, quotient and power — which turn multiplication into addition and make logarithmic equations solvable.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.4 Evaluate Logarithms and Graph Logarithmic Functions §7.4, pp. 498-505 — everything on these slides traces back here
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