Math Core

An exponential function answers "what do I get if I raise $b$ to this power?" The reverse question, "what power of $b$ gives this number?", is answered by a logarithm. Logarithms let you solve for exponents, and they turn quantities that span enormous ranges (acidity, loudness, earthquake energy) into manageable numbers.

Logarithms as exponents

Since $y = b^x$ is one-to-one, it has an inverse function. That inverse is the logarithm with base $b$ .

Definition

Logarithm

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

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

Read $\log_b x$ as "log base $b$ of $x$ ." It is the exponent you put on $b$ to get $x$ .

So every logarithm statement is an exponential statement in disguise:

Logarithmic form	Exponential form
$\log_2 32 = 5$	$2^5 = 32$
$\log_{10} 0.001 = -3$	$10^{-3} = 0.001$
$\log_9 3 = \frac{1}{2}$	$9^{1/2} = 3$
$\log_b 1 = 0$	$b^0 = 1$

In both forms the base stays the base. The logarithm is the exponent.

Two bases get their own notation because calculators have keys for them:

The common logarithm $\log x$ means $\log_{10} x$ .
The natural logarithm $\ln x$ means $\log_e x$ .

Evaluating logarithms

To evaluate $\log_b x$ by hand, write $x$ as a power of $b$ .

Worked example: Evaluate without a calculator

Evaluate (a) $\log_3 81$ , (b) $\log_5 \dfrac{1}{25}$ , (c) $\log_{8} 4$ , (d) $\ln e^{7}$ .

Solution.

(a) $81 = 3^4$ , so $\log_3 81 = 4$ .

(b) $\dfrac{1}{25} = 5^{-2}$ , so $\log_5 \dfrac{1}{25} = -2$ .

(c) Write both numbers as powers of $2$ : $8 = 2^3$ and $4 = 2^2$ . You need $8^y = 4$ , so $2^{3y} = 2^2$ , giving $3y = 2$ and $\log_8 4 = \dfrac{2}{3}$ .

(d) $\ln e^7$ asks what power of $e$ gives $e^7$ . The answer is $7$ .

Part (d) is an instance of the inverse relationship. Because $b^x$ and $\log_b x$ undo each other:

Inverse properties

For $b > 0$ , $b \ne 1$ :

\log_b\left(b^{x}\right) = x \text{ for every real } x, \qquad b^{\log_b x} = x \text{ for every } x > 0.

In particular, $\log_b 1 = 0$ , $\log_b b = 1$ , $\ln e^x = x$ and $e^{\ln x} = x$ .

For instance, $10^{\log 6} = 6$ and $e^{\ln 2.5} = 2.5$ without any computation.

Common mistake

You cannot take the log of $0$ or of a negative number. No power of a positive base equals $0$ or a negative number, so expressions like $\log_2(-8)$ and $\ln 0$ are undefined. The output of a log can be negative, though: $\log_{10} 0.01 = -2$ is perfectly fine.

Graphs of logarithmic functions

The graph of an inverse function is the reflection of the original graph across the line $y = x$ . Every point $(a, c)$ on $y = 2^x$ becomes $(c, a)$ on $y = \log_2 x$ . For example, $(3, 8)$ becomes $(8, 3)$ .

y = log₂ x is the reflection of y = 2ˣ across y = x.Open in grapher →

Swapping $x$ and $y$ swaps every feature: the domain and range trade places, and the horizontal asymptote becomes a vertical one.

Features of y = log_b x

Domain: $x > 0$ . Range: all real numbers.
$x$ -intercept: $(1, 0)$ . There is no $y$ -intercept.
Vertical asymptote: $x = 0$ .
Increasing if $b > 1$ ; decreasing if $0 < b < 1$ .
The graph also passes through $(b, 1)$ .

Logarithms grow extremely slowly. For $y = \log x$ to reach $6$ , $x$ must reach $1{,}000{,}000$ .

y = ln x (base e) and y = log x (base 10). Both cross the x-axis at x = 1.Open in grapher →

Transformations and domain

For $f(x) = a\log_b(x - h) + k$ , the vertical asymptote moves with the horizontal shift to $x = h$ , and the domain is $x > h$ . More generally, the domain of any logarithmic function is found by requiring the argument to be positive.

Worked example: Domain, asymptote and intercept

Let $f(x) = \ln(2x - 6) + 1$ . Find the domain, the vertical asymptote and the $x$ -intercept.

Solution. The argument must be positive: $2x - 6 > 0$ , so $x > 3$ . The domain is $x > 3$ and the vertical asymptote is $x = 3$ .

For the $x$ -intercept, set $f(x) = 0$ :

\ln(2x - 6) = -1 \quad\Longrightarrow\quad 2x - 6 = e^{-1} \quad\Longrightarrow\quad x = \frac{6 + e^{-1}}{2} \approx 3.184.

The intercept is just to the right of the asymptote, as the graph suggests.

y = ln(2x - 6) + 1x = 3Open in grapher →

Tip

To convert a log equation, use the "spiral": in $\log_b x = y$ , start at the base $b$ , go to the other side for the exponent $y$ , and come back for $x$ . That reads $b^y = x$ .

Logarithmic scales

When a quantity varies over many powers of ten, it is often reported on a logarithmic scale, where each step of $1$ means a factor of $10$ .

Acidity: $\text{pH} = -\log\left[\text{H}^+\right]$ , where $\left[\text{H}^+\right]$ is the hydrogen-ion concentration in moles per liter.
Loudness: $L = 10\log\dfrac{I}{I_0}$ decibels, where $I_0 = 10^{-12}$ watts per square meter is the quietest sound a person can hear.
Earthquakes: each whole-number step on the magnitude scale means $10$ times the ground motion.

Worked example: pH and decibels

(a) A sample of tomato juice has $\left[\text{H}^+\right] = 6.3 \times 10^{-5}$ . Find its pH.

(b) One sound is $1000$ times as intense as another. How many decibels louder is it?

Solution.

(a) $\text{pH} = -\log\left(6.3 \times 10^{-5}\right) \approx -(-4.20) = 4.20$ . The juice is acidic (below $7$ ).

(b) If $I_2 = 1000I_1$ , then

L_2 = 10\log\frac{1000I_1}{I_0} = 10\left(\log 1000 + \log\frac{I_1}{I_0}\right) = 30 + L_1.

It is $30$ decibels louder. Multiplying intensity by $1000 = 10^3$ adds $3$ steps of $10$ dB.

Part (b) previews the next lesson: the log of a product is the sum of the logs. That property is exactly why log scales turn multiplication into addition.