2.12 - Logarithmic Function Manipulation

Overview

Logs follow a few rules that let you rewrite them: the product property, the power property, and the change of base property. Each one comes straight from an exponent rule, and each one also says something about graphs. On the AP exam you'll use them to rewrite expressions, to solve equations (Topic 2.13), and to see when one log function is a transformation of another.

Where the properties come from

A log is an exponent:

\log_b x

is the power you raise

b

to in order to get

x

. So every exponent rule turns into a log property:

Exponent rule	Log property
$b^m \cdot b^n = b^{m+n}$	$\log_b(xy) = \log_b x + \log_b y$
$\frac{b^m}{b^n} = b^{m-n}$	$\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$
$\left(b^m\right)^n = b^{mn}$	$\log_b\left(x^n\right) = n\log_b x$

All of these need

x > 0

and

y > 0

, since you can only take logs of positive numbers.

The product property

\log_b(xy) = \log_b x + \log_b y

Why it works: let

m = \log_b x

and

n = \log_b y

, so

x = b^m

and

y = b^n

. Then

xy = b^m \cdot b^n = b^{m+n}

, which means

\log_b(xy) = m + n = \log_b x + \log_b y

Examples:

\log_2 24 = \log_2(8 \cdot 3) = \log_2 8 + \log_2 3 = 3 + \log_2 3

. Going the other way,

\log 4 + \log 25 = \log 100 = 2

What it means for graphs

Take

f(x) = \log_b(kx)

with

k > 0

. Multiplying the input by

k

is a horizontal dilation by a factor of

\frac{1}{k}

. But the product property splits it into a sum:

f(x) = \log_b(kx) = \log_b k + \log_b x = a + \log_b x, \quad \text{where } a = \log_b k

So every horizontal dilation of a log function is the same as a vertical translation. For example,

\log_2(4x) = \log_2 4 + \log_2 x = 2 + \log_2 x

. Squeezing

y = \log_2 x

toward the

y

-axis by a factor of

\frac{1}{4}

gives exactly the same graph as shifting it up 2.

2.12 - Logarithmic Function Manipulation

In the graph, the point

(8, 3)

y = \log_2 x

moves to

(2, 3)

when the inputs are multiplied by

\frac{1}{4}

, and the point

(2, 1)

moves to

(2, 3)

when you shift up 2. Both moves land on the same red curve.

The quotient property

You'll also use a quotient property. It comes from the product and power properties, since dividing by

y

is multiplying by

y^{-1}

\log_b\left(\frac{x}{y}\right) = \log_b x + \log_b\left(y^{-1}\right) = \log_b x - \log_b y

Example:

\log_3\left(\frac{x}{9}\right) = \log_3 x - \log_3 9 = \log_3 x - 2

. On a graph, dividing the input by 9 (a horizontal stretch by a factor of 9) is the same as shifting down 2.

The power property

\log_b\left(x^n\right) = n\log_b x

Why it works: if

m = \log_b x

, then

x = b^m

, so

x^n = \left(b^m\right)^n = b^{mn}

. That means

\log_b\left(x^n\right) = mn = n\log_b x

Examples:

\log_5 125 = \log_5 5^3 = 3

\log_2\sqrt{x} = \log_2 x^{1/2} = \frac{1}{2}\log_2 x

\ln\left(\frac{1}{x}\right) = \ln x^{-1} = -\ln x

What it means for graphs

Raising the input to a power,

f(x) = \log_b\left(x^k\right)

, gives

f(x) = k\log_b x

. That's a vertical dilation by a factor of

k

. So

y = \log_3\left(x^4\right)

y = \log_3 x

stretched vertically by a factor of 4: every output is 4 times as big.

Domain warning:

\log_3\left(x^4\right)

is defined for every

x \neq 0

, since

x^4 > 0

, but

4\log_3 x

only works for

x > 0

. The two match for

x > 0

. This difference is one way extraneous solutions show up in Topic 2.13.

The change of base property

\log_b x = \frac{\log_a x}{\log_a b}, \quad \text{where } a > 0 \text{ and } a \neq 1

Why it works: let

y = \log_b x

, so

b^y = x

. Take

\log_a

of both sides:

\log_a\left(b^y\right) = \log_a x

. The power property turns this into

y\log_a b = \log_a x

, so

y = \frac{\log_a x}{\log_a b}

Use 1, calculators. Most calculators only have

\log

(base 10) and

\ln

(base

e

). To find

\log_3 50

\log_3 50 = \frac{\ln 50}{\ln 3} \approx \frac{3.9120}{1.0986} \approx 3.561

That makes sense:

3^3 = 27

and

3^4 = 81

, so the answer should be between 3 and 4. (Check:

3^{3.561} \approx 50

Use 2, graphs. The denominator

\log_a b

is just a number, so

\log_b x = \frac{1}{\log_a b} \cdot \log_a x

. That means all logarithmic functions are vertical dilations of each other. For example, since

\log_2 8 = 3

\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{1}{3}\log_2 x

So the graph of

y = \log_8 x

is the graph of

y = \log_2 x

compressed vertically by a factor of

\frac{1}{3}

All three curves pass through

(1, 0)

, and each one reaches a height of 1 at its own base:

(2, 1)

(e, 1)

, and

(10, 1)

. Each is a vertical dilation of the others. For example,

\log_2 x = \frac{\ln x}{\ln 2} \approx 1.443\ln x

The natural logarithm

The natural logarithm is the log with base

e \approx 2.718

\ln x = \log_e x

. It's a logarithmic function like any other, so every property above works for it.

$\ln 1 = 0$ and $\ln e = 1$

$\ln\left(e^x\right) = x$ for every $x$ , and $e^{\ln x} = x$ for $x > 0$ , because $\ln x$ and $e^x$ are inverses

$\ln(5x) = \ln 5 + \ln x$ , $\ln\left(x^3\right) = 3\ln x$ , and $\log_b x = \frac{\ln x}{\ln b}$

You'll see

\ln

a lot in modeling (Topic 2.14) and in calculus, because

e

shows up naturally in continuous growth.

Summary

Property	Rule	What it means for a graph
Product	$\log_b(xy) = \log_b x + \log_b y$	a horizontal dilation is the same as a vertical translation
Quotient	$\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$	dividing the input is also a vertical translation
Power	$\log_b\left(x^n\right) = n\log_b x$	raising the input to a power is a vertical dilation
Change of base	$\log_b x = \frac{\log_a x}{\log_a b}$	every log function is a vertical dilation of every other

Examples

Example 1: Expanding a log

Write

\log_2\left(\frac{8x^3}{y}\right)

as a sum and difference of simpler logs. Assume

x > 0

and

y > 0

Split the quotient, then the product, then use the power property:

\log_2\left(\frac{8x^3}{y}\right) = \log_2 8 + \log_2\left(x^3\right) - \log_2 y

= 3 + 3\log_2 x - \log_2 y

Answer:

3 + 3\log_2 x - \log_2 y

Example 2: Condensing into one log

Write

2\ln x + \ln(x + 1) - \ln 5

as a single logarithm.

Use the power property first, so every log has a coefficient of 1. Then combine with the product and quotient properties:

2\ln x + \ln(x + 1) - \ln 5 = \ln\left(x^2\right) + \ln(x + 1) - \ln 5

= \ln\left(\frac{x^2(x + 1)}{5}\right)

Answer:

\ln\left(\frac{x^2(x + 1)}{5}\right)

, for

x > 0

Example 3: Log functions as transformations of each other (AP-style)

Let

f(x) = \log_2\left(8x^5\right)

g(x) = \log_2 x

, and

h(x) = \log_4 x

Write $f(x)$ in the form $f(x) = c + k\log_2 x$ .

Describe how the graph of $f$ is related to the graph of $g$ .

Write $h$ as a transformation of $g$ .

Part 1: use the product property, then the power property:

f(x) = \log_2 8 + \log_2\left(x^5\right) = 3 + 5\log_2 x

Part 2: the graph of

f

is the graph of

g

dilated vertically by a factor of 5, then translated up 3. (Check with

x = 2

f(2) = \log_2 256 = 8

, and

3 + 5\log_2 2 = 8

Part 3: use change of base with base 2:

h(x) = \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x = \frac{1}{2}g(x)

Answer:

f(x) = 3 + 5\log_2 x

, a vertical dilation of

g

by 5 and a shift up 3. The graph of

h

is the graph of

g

dilated vertically by a factor of

\frac{1}{2}

. (Check:

h(16) = 2

and

\frac{1}{2}\log_2 16 = 2

Common mistakes

$\log_b(x + y)$ is not $\log_b x + \log_b y$ . The product property is about the log of a product, not a sum. $\log_b(x + y)$ can't be split.

$\left(\log_b x\right)^2$ is not $2\log_b x$ . The power property only works when the power is on the input: $\log_b\left(x^2\right) = 2\log_b x$ .

$\frac{\log_b x}{\log_b y}$ is not $\log_b\left(\frac{x}{y}\right)$ . A quotient of logs is a change of base: $\frac{\log_b x}{\log_b y} = \log_y x$ .

Flipping change of base. The input goes on top and the base on the bottom: $\log_3 50 = \frac{\ln 50}{\ln 3}$ , not $\frac{\ln 3}{\ln 50}$ .

Combining before using the power property. In $2\ln x + \ln(x + 1)$ , turn $2\ln x$ into $\ln\left(x^2\right)$ first. The 2 does not multiply $(x + 1)$ .

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