Math Core

You can answer "what is $2^5$ ?" by multiplying. But how do you answer " $2$ to what power is $32$ ?" or, harder, " $2$ to what power is $50$ ?" Questions like these come up whenever you solve for a time in a growth model. A logarithm is the tool that answers them: it undoes an exponential the way a square root undoes a square.

What a logarithm is

A logarithm is an exponent. The expression $\log_2 32$ asks, "What power of $2$ gives $32$ ?" Since $2^5 = 32$ , the answer is $5$ .

Definition

Logarithm

For $b > 0$ , $b \ne 1$ and $x > 0$ ,

\log_b x = y \quad\text{means}\quad b^y = x.

Read $\log_b x$ as "log base $b$ of $x$ ." The base of the logarithm is the base of the power, and the logarithm itself is the exponent.

Every logarithmic statement is an exponential statement in disguise. Being able to switch between the two forms is the single most useful skill in this unit.

Worked example: Switching forms

Rewrite each equation in the other form.

$2^5 = 32$
$\log_3 \dfrac{1}{9} = -2$
$10^{-3} = 0.001$
$\log_b 7 = x$

Solution. The base stays the base, and the exponent is what the log equals.

$\log_2 32 = 5$
$3^{-2} = \dfrac{1}{9}$
$\log_{10} 0.001 = -3$
$b^x = 7$

Evaluating logarithms

To evaluate $\log_b x$ by hand, write $x$ as a power of $b$ . It often helps to set the log equal to $y$ and solve $b^y = x$ .

Worked example: Evaluate without a calculator

$\log_4 64$
$\log_5 \dfrac{1}{25}$
$\log_9 3$
$\log_8 4$

Solution.

$4^3 = 64$ , so $\log_4 64 = 3$ .
$\dfrac{1}{25} = 5^{-2}$ , so $\log_5 \dfrac{1}{25} = -2$ .
$3 = \sqrt{9} = 9^{1/2}$ , so $\log_9 3 = \dfrac{1}{2}$ .
Let $\log_8 4 = y$ , so $8^y = 4$ . Write both sides as powers of $2$ : $2^{3y} = 2^2$ . Then $3y = 2$ and $y = \dfrac{2}{3}$ .

A few values follow straight from the definition and are worth memorizing. For any valid base $b$ :

\log_b 1 = 0 \qquad \log_b b = 1 \qquad \log_b b^x = x \qquad b^{\log_b x} = x

The first holds because $b^0 = 1$ and the second because $b^1 = b$ . The last two say that "raise $b$ to a power" and "take $\log_b$ " cancel each other.

Common and natural logarithms

Two bases get special notation because they are used so often.

The common logarithm has base $10$ and is written $\log x$ with no base shown. So $\log 1000 = 3$ .
The natural logarithm has base $e$ and is written $\ln x$ . So $\ln e^4 = 4$ and $\ln 1 = 0$ .

Your calculator has keys for both. For example, $\log 50 \approx 1.699$ (because $10^{1.699} \approx 50$ ) and $\ln 20 \approx 2.996$ (because $e^{2.996} \approx 20$ ). You'll learn in the next lesson how to use these keys to find logs in any base.

Logarithmic functions and their graphs

Since $\log_b x$ undoes $b^x$ , the function $g(x) = \log_b x$ is the inverse of $f(x) = b^x$ . Its graph is the reflection of the exponential graph across the line $y = x$ : every point $(a, c)$ on $y = 2^x$ becomes $(c, a)$ on $y = \log_2 x$ .

y = 2^x and y = log₂ x are reflections of each other across the dashed line y = x.Open in grapher →

Swapping $x$ and $y$ swaps all the features:

	$y = b^x$	$y = \log_b x$
Domain	all real numbers	$x > 0$
Range	$y > 0$	all real numbers
Key point	$(0, 1)$	$(1, 0)$
Asymptote	horizontal, $y = 0$	vertical, $x = 0$

You can't take the log of zero or a negative number

Because $b^y$ is always positive, $\log_b x$ is defined only for $x > 0$ . For a transformed log like $y = \log_b(x - h) + k$ , the domain is $x > h$ and the vertical asymptote is $x = h$ .

Worked example: A transformed logarithm

Describe $y = \log_2(x - 3) + 1$ : its domain, asymptote, and one exact point.

Solution. The argument must be positive: $x - 3 > 0$ , so the domain is $x > 3$ and the vertical asymptote is $x = 3$ . The parent graph passes through $(1, 0)$ and $(2, 1)$ ; shifting right $3$ and up $1$ gives $(4, 1)$ and $(5, 2)$ .

Check $(5, 2)$ : $\log_2(5 - 3) + 1 = \log_2 2 + 1 = 1 + 1 = 2$ . ✓

y = log₂(x − 3) + 1 with its vertical asymptote x = 3.Open in grapher →

Common mistake

$\log_2 8$ is not $\dfrac{8}{2}$ or $8^2$ . It's the exponent that turns $2$ into $8$ , which is $3$ . When in doubt, rewrite the log as an exponential equation.