Unit 2 · Topic 2.12 · about 30 minutes · 7 parts · 13 scored checks, 1 of them multiple choice

Logarithmic Function Manipulation

Built from College Board's course framework · Every answer key independently checked

Part 1 Think first

Squeezed or lifted?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

Ana and Ben graph $y = \log_{2} (8 x)$ on the same screen as $y = \log_{2} x$ . Ana says the new curve is the old one squeezed toward the $y$ -axis. Ben says it is the old one moved straight up. Who is right?

In short. Manipulating a logarithmic function means rewriting it in an equivalent form with the product, power and change of base properties, and reading what the new form says about the graph.

Product property: $\log_{b} (x y) = \log_{b} x + \log_{b} y$ . So a horizontal dilation $\log_{b} (k x)$ is the vertical translation $\log_{b} k + \log_{b} x$ .
Power property: for $x > 0$ , $\log_{b} x^{n} = n \log_{b} x$ . So $\log_{b} x^{k}$ is a vertical dilation of $\log_{b} x$ by a factor of $k$ .
Change of base: $\log_{b} x = \frac{\log_{a} x}{\log_{a} b}$ , so every logarithmic function is a vertical dilation of every other. The natural logarithm is $\ln x = \log_{e} x$ .

Part 2 Learn it

Three properties, each checked as you go

2a. The product property

A logarithm is an exponent, so it follows the rules exponents follow. Multiply two powers of the same base and the exponents add: $2^{3} \cdot 2^{2} = 2^{5}$ . In logarithm form, that is the product property:

\log_{b} (x y) = \log_{b} x + \log_{b} y

for positive $x$ and $y$ . Check it: $\log_{2} (8 \cdot 4) = \log_{2} 32 = 5$ , and $\log_{2} 8 + \log_{2} 4 = 3 + 2 = 5$ .

Now watch what it does to a graph. In $g (x) = \log_{2} (4 x)$ every input is multiplied by 4 before the logarithm acts, so $g$ is a horizontal dilation of $f (x) = \log_{2} x$ , squeezed toward the $y$ -axis. The product property rewrites it:

\log_{2} (4 x) = \log_{2} 4 + \log_{2} x = 2 + \log_{2} x

So the same curve is $f$ moved up 2. The left panel of the diagram shows both moves landing on one curve. In general, $\log_{b} (k x) = a + \log_{b} x$ with $a = \log_{b} k$ . When $0 < k < 1$ , $a$ is negative and the shift is down: $\log_{2} (\frac{x}{8}) = - 3 + \log_{2} x$ .

The property needs a product inside the logarithm. A sum does not split: $\log_{2} (4 + 4) = \log_{2} 8 = 3$ , but $\log_{2} 4 + \log_{2} 4 = 4$ .

Quick check · Fill the statement

No calculator. Rewrite $g (x) = \log_{3} (9 x)$ with the product property, then describe its graph. Tap a blank, then tap an answer.

The graph of $g (x) = \log_{3} (9 x)$ is the graph of $f (x) = \log_{3} x$ by .

2b. The power property

Raise a power to a power and the exponents multiply: $(2^{3})^{2} = 2^{6}$ . In logarithm form, that is the power property, for $x > 0$ :

\log_{b} x^{n} = n \log_{b} x

Check it: $\log_{2} 8^{2} = \log_{2} 64 = 6$ , and $2 \log_{2} 8 = 2 \cdot 3 = 6$ . The power can be a fraction, so a root comes out front too: $\log_{b} \sqrt{x} = \log_{b} x^{\frac{1}{2}} = \frac{1}{2} \log_{b} x$ .

On a graph, $p (x) = \log_{2} x^{3}$ cubes each input before taking the logarithm. Rewritten, $p (x) = 3 \log_{2} x$ , a vertical dilation of $\log_{2} x$ by a factor of 3: every output is tripled. At $x = 4$ , $\log_{2} x$ is 2 and $p$ is 6. Read both forms for $x > 0$ , where $\log_{2} x$ is defined. The right panel of the diagram shows the same idea with a power of 2.

The exponent has to sit on the input. $(\log_{2} 8)^{2}$ is $3^{2} = 9$ , not 6, because there the whole logarithm is squared. In $\log_{2} (5 x^{3})$ the 3 belongs to $x$ alone, so split the product first: $\log_{2} 5 + 3 \log_{2} x$ .

The two properties also cover a quotient, since dividing by $y$ is multiplying by $y^{- 1}$ : $\log_{b} \frac{x}{y} = \log_{b} x + (- 1) \log_{b} y = \log_{b} x - \log_{b} y$ .

Quick check · Spot the error

No calculator. A student rewrites $\log_{2} (8 x^{3})$ and then evaluates it at $x = 4$ . Tap the line where the mistake is made.

2c. Change of base, and the natural logarithm

Your calculator has two logarithm keys. LOG is base 10. LN is the natural logarithm, the logarithm with the natural base $e \approx 2.718$ : $\ln x = \log_{e} x$ . So $\ln e = 1$ and $\ln e^{4} = 4$ , for the same reason that $\log 10^{4} = 4$ .

For any other base you switch to one you have. The change of base property says

\log_{b} x = \frac{\log_{a} x}{\log_{a} b}

for any base $a > 0$ with $a \neq 1$ . So $\log_{5} 40 = \frac{\ln 40}{\ln 5} \approx 2.292$ , and $\frac{\log 40}{\log 5}$ gives the same number. The base you started with goes in the denominator.

Here is why it works. Let $y = \log_{b} x$ , so $b^{y} = x$ . Take $\log_{a}$ of both sides and use the power property: $y \log_{a} b = \log_{a} x$ . Divide by $\log_{a} b$ .

That denominator is just a number, so $\log_{b} x$ is a constant times $\log_{a} x$ . For example, $\log_{8} x = \frac{\log_{2} x}{\log_{2} 8} = \frac{1}{3} \log_{2} x$ . Every logarithmic function is a vertical dilation of every other one, and the graph of $\log_{8} x$ is the graph of $\log_{2} x$ with every output divided by 3.

Quick check · Match

No calculator. Each function is a vertical dilation of $\log_{2} x$ . Tap a function, then the rewrite that equals it.

Part 3 Watch it done

One rewrite, all three properties

Watch each step, then do the last one yourself.

Rewrite $h (x) = \log_{4} (16 x^{5})$ in the form $a + c \log_{2} x$ , for $x > 0$ , and say what the rewrite tells you about the graph of $h$ .

Step 1 · Split the product
By the product property, $h (x) = \log_{4} 16 + \log_{4} x^{5}$ .
Step 2 · Evaluate, then bring the power down
$\log_{4} 16 = 2$ , because $4^{2} = 16$ . By the power property, $\log_{4} x^{5} = 5 \log_{4} x$ . So $h (x) = 2 + 5 \log_{4} x$ .
Why this step?
Why split the product before bringing the 5 to the front?
Because the 5 is an exponent on $x$ alone, not on $16 x$ . Writing $5 \log_{4} (16 x)$ would raise the 16 to the fifth power as well, which changes the function.
Step 3 · Change the base to 2
By the change of base property, $\log_{4} x = \frac{\log_{2} x}{\log_{2} 4} = \frac{\log_{2} x}{2}$ . Substitute that into $2 + 5 \log_{4} x$ .
Your step · Calculate

No calculator. Finish it. With $h (x) = 2 + c \log_{2} x$ , what is $c$ ? Give a decimal.

Your answer

So $h (x) = 2 + 2.5 \log_{2} x$ . The original form hid both numbers. The rewrite shows them: stretch $\log_{2} x$ vertically by 2.5, then lift it 2.

Part 4 Try it

Make two graphs land on each other

The grapher below has one logarithm, $\log_{2} x$ , and four sliders. That is enough to test both graph claims in this lesson, because a horizontal dilation and a vertical translation should give the same curve, and so should a change of base and a vertical dilation.

Lab

Function Grapher

Choose the logarithmic family. The lab draws $g (x) = a \log_{2} (b (x + c)) + d$ . Keep $c = 0$ the whole time. Work each answer out with a property first, then move the sliders and check that the colored curve lands where you said it would.

Open the full Function Grapher lab

Try it · Calculate

No calculator. In the grapher, set $a = 1$ , $b = 0.25$ and $d = 0$ , so the curve is $y = \log_{2} (0.25 x)$ . Then set $b$ back to 1. What value of $d$ draws the same curve?

Your answer

Try it · Decide and justify

No calculator. In the grapher, set $b = 1$ and $d = 0$ . Which value of $a$ makes the curve $y = \log_{16} x$ ?

Which reason supports your answer?

Practice

Practice until it is automatic

Sums and multiples of logarithms with exact values, new every time. Combine them into one logarithm with the product or power property, then read off the exponent.

Logarithm values practice page

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 37.5% of the exam score. This one is shaped like Question 4, Symbolic Manipulations.

Free-response practice · Question 4 style

No calculator. (A) The function $f$ is given by $f (x) = \ln (e^{3} x^{2})$ for $x > 0$ . Rewrite $f (x)$ in the form $a + c \ln x$ , where $a$ and $c$ are constants. (B) The function $k$ is given by $k (x) = \log_{9} x$ . Rewrite $k (x)$ as a constant times $\log_{3} x$ , and describe how the graph of $k$ is related to the graph of $y = \log_{3} x$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(A) By the product property, $f (x) = \ln e^{3} + \ln x^{2}$ . Since $\ln$ is the logarithm with base $e$ , $\ln e^{3} = 3$ . By the power property, $\ln x^{2} = 2 \ln x$ for $x > 0$ . So $f (x) = 3 + 2 \ln x$ . (B) By the change of base property, $k (x) = \frac{\log_{3} x}{\log_{3} 9} = \frac{\log_{3} x}{2} = \frac{1}{2} \log_{3} x$ . The graph of $k$ is a vertical dilation of the graph of $y = \log_{3} x$ by a factor of $\frac{1}{2}$ : every output is half as large.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Uses the product property to write $f (x) = \ln e^{3} + \ln x^{2}$ .
Gets $f (x) = 3 + 2 \ln x$ , using $\ln e^{3} = 3$ and the power property.
Writes $k (x) = \frac{\log_{3} x}{\log_{3} 9} = \frac{1}{2} \log_{3} x$ by change of base.
Describes the graph of $k$ as a vertical dilation of $y = \log_{3} x$ by a factor of $\frac{1}{2}$ .

Part 6 Check yourself

Three values, many more

The three case questions use the same table. Answer them from it.

The table gives three values of a logarithmic function $q (x) = \log_{b} x$ , rounded to three decimal places. The base $b$ is greater than 1, and you do not need to know it.

Selected values of $q (x) = \log_{b} x$
$x$	$q (x)$
2	0.356
3	0.565
5	0.827

Case question 1 · Calculate

No calculator. Use the table of $q (x) = \log_{b} x$ to find $q (18)$ . Give a decimal.

Your answer

Case question 2 · Decide and justify

No calculator. Dev says that for $q (x) = \log_{b} x$ , $q (5) = q (2) + q (3)$ , because $5 = 2 + 3$ . Is he right?

Which reason supports your answer?

Case question 3 · Multiple choice

No calculator. For $q (x) = \log_{b} x$ , which expression is equal to $\log_{2} 5$ ?

The rest stand on their own and do not use the case.

Check yourself · Put it in order

No calculator. Put the lines of this rewrite of $\log_{2} \sqrt{32 x}$ in order, for $x > 0$ . Each line takes one step from the line before it, using one property or simplifying. Tap them from first to last.

Check yourself · Sort it

No calculator. Rewrite each function with a property of logarithms. Is its graph, for $x > 0$ , a vertical translation of $y = \log_{2} x$ , a vertical dilation of it, or neither? Tap a card, then a bin.

Vertical translation

Vertical dilation

Neither

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 2.10 Inverses of Exponential Functions · Spot the error

No calculator. A student sketches $f (x) = \log_{3} x$ from the graph of $g (x) = 3^{x}$ . Tap the line with the mistake.

From 2.9 Logarithmic Expressions · Decide and justify

No calculator. A classmate says $\log_{2} 8$ and $\log_{8} 2$ are equal, because both use the numbers 2 and 8. Is he right?

Which reason supports your answer?

In three lines

Product property: $\log_{b} (x y) = \log_{b} x + \log_{b} y$ , so $\log_{b} (k x)$ is $\log_{b} x$ translated vertically by $\log_{b} k$ .
Power property: for $x > 0$ , $\log_{b} x^{n} = n \log_{b} x$ , so $\log_{b} x^{k}$ is $\log_{b} x$ dilated vertically by $k$ .
Change of base: $\log_{b} x = \frac{\log_{a} x}{\log_{a} b}$ , so all logarithmic functions are vertical dilations of each other, and $\ln x = \log_{e} x$ .

Vocabulary

Product property for logarithms: $\log_{b} (x y) = \log_{b} x + \log_{b} y$ , which makes the horizontal dilation $\log_{b} (k x)$ the same as the vertical translation $\log_{b} k + \log_{b} x$ .
Power property for logarithms: For $x > 0$ , $\log_{b} x^{n} = n \log_{b} x$ , so raising the input to the power $k$ dilates the graph vertically by a factor of $k$ .
Change of base property for logarithms: $\log_{b} x = \frac{\log_{a} x}{\log_{a} b}$ for any base $a > 0$ with $a \neq 1$ , which makes every logarithmic function a vertical dilation of every other.
Natural base e: The number $e \approx 2.718$ ; the logarithmic function with this base is $\ln x = \log_{e} x$ , on the LN key of a calculator.

Course alignment, for teachers

AP Precalculus topic 2.12, Unit 2: Exponential and Logarithmic Functions.

Skill 1.B: Express functions, equations, or expressions in analytically equivalent forms that are useful in a given mathematical or applied context.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 4 essential knowledge statements of the topic to the part that teaches each one: 4 are checked by a scored question. The map is in the lesson file.

Squeezed or lifted?

Three properties, each checked as you go

2a. The product property

2b. The power property

2c. Change of base, and the natural logarithm

One rewrite, all three properties

Make two graphs land on each other

Function Grapher

Practice until it is automatic

Write it the way the exam asks

Three values, many more

Bring back earlier ideas

Vocabulary

Labs that use this topic