Logarithms as Inverse Functions, foundations for calculus lesson

If money doubles every year, how many years until it is eight times as large? You know the output of the power and want the exponent. A logarithm answers exactly that question, and it does so as a function in its own right, the inverse of an exponential function.

Key terms

Logarithm, base and argument

Definition Let $a > 0$ with $a \ne 1$ , and let $x > 0$ . The logarithm of $x$ with base $a$ , written $\log_a x$ , is the unique real number $y$ such that $a^y = x$ . In symbols,

\begin{gathered} \log_a x = y \\[0.4em] \text{exactly when} \quad a^y = x. \end{gathered}

The number $a$ is the base and the number $x$ is the argument. Uniqueness holds because $a^y$ is strictly increasing or strictly decreasing in $y$ , and existence holds because $a^y$ takes every positive value.

In plain words The logarithm $\log_a x$ asks which power of $a$ gives $x$ . The base is the number being raised to a power and the argument is the result you want to reach.

Example Since $2^3 = 8$ , we have $\log_2 8 = 3$ , so money that doubles yearly becomes eight times as large in $3$ years. Also $\log_{10} 1000 = 3$ and $\log_3 \frac{1}{9} = -2$ because $3^{-2} = \frac{1}{9}$ .

Exponent as output

Definition The equations $\log_a x = y$ and $a^y = x$ are two forms of the same statement, the logarithmic form and the exponential form. In the logarithmic form the exponent $y$ is the output of the function and the power $x$ is the input.

In plain words An exponential function takes an exponent and returns a power. A logarithm takes the power and returns the exponent.

Example Each fact can be written both ways.

exponential form	logarithmic form
$5^2 = 25$	$\log_5 25 = 2$
$4^{1/2} = 2$	$\log_4 2 = \frac{1}{2}$
$10^{-3} = 0.001$	$\log_{10} 0.001 = -3$
$\left(\frac{1}{2}\right)^{-3} = 8$	$\log_{1/2} 8 = -3$

Inverse relationship

Definition For $a > 0$ with $a \ne 1$ , the function $f(x) = a^x$ is one-to-one from $\mathbb{R}$ onto $(0, \infty)$ , and its inverse is $f^{-1}(x) = \log_a x$ from $(0, \infty)$ onto $\mathbb{R}$ . The two cancellation identities are

\begin{gathered} \log_a\!\left(a^y\right) = y \ \text{ for every real } y, \\[0.4em] a^{\log_a x} = x \ \text{ for every } x > 0. \end{gathered}

In particular $\log_a 1 = 0$ and $\log_a a = 1$ .

In plain words Raising $a$ to a power and taking the base $a$ logarithm undo each other, as long as you stay inside their domains.

Example The identities give $\log_7\!\left(7^{4.2}\right) = 4.2$ and $3^{\log_3 20} = 20$ . The expression $2^{\log_2(-5)}$ has no real value, because $\log_2(-5)$ does not exist.

Positive argument, domain and range

Definition The real function $\log_a x$ has domain $(0, \infty)$ and range $\mathbb{R}$ . Every real output $y$ is attained, at the positive input $x = a^y$ . Zero and negative arguments have no real logarithm, because $a^y > 0$ for every real $y$ . For a composite such as $\log_a(g(x))$ the domain is the set of $x$ in the domain of $g$ with $g(x) > 0$ .

In plain words A positive base raised to any power is positive, so only positive numbers can be reached. In return, every exponent, large, small or negative, is the logarithm of some positive number.

Example The value $\log_2(-8)$ is undefined, since no real $y$ gives $2^y = -8$ . The function $\log_2(6 - 2x)$ needs $6 - 2x > 0$ , which means $x < 3$ , so its domain is $(-\infty, 3)$ .

Logarithmic graph

Definition The graph of $y = \log_a x$ is the reflection of the graph of $y = a^x$ across the line $y = x$ . A point $(p, q)$ lies on the graph of $a^x$ exactly when $(q, p)$ lies on the graph of $\log_a x$ . Every logarithmic graph passes through $(1, 0)$ and $(a, 1)$ .

In plain words Take a table of powers, swap the two rows, and plot. The new points trace the logarithm.

Example For base $2$ , swapping the rows of a power table gives

$x$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$	$8$
$\log_2 x$	$-2$	$-1$	$0$	$1$	$2$	$3$

The graph rises slowly to the right, crosses the $x$ -axis at $(1, 0)$ and drops steeply as $x$ approaches $0$ .

Vertical asymptote

Definition For $a > 1$ , $\log_a x \to -\infty$ as $x \to 0^+$ , and for $0 < a < 1$ , $\log_a x \to \infty$ as $x \to 0^+$ . In both cases the line $x = 0$ is a vertical asymptote of the graph. The graph lies entirely to the right of it and never touches it, since $0$ is not in the domain. This is the horizontal asymptote $y = 0$ of $a^x$ reflected across $y = x$ .

In plain words As the argument gets close to $0$ , you need a huge negative exponent (or a huge positive one when the base is below $1$ ) to reach it, so the graph runs off along the $y$ -axis.

Example For base $2$ , the input $2^{-20} \approx 0.00000095$ has logarithm $\log_2\!\left(2^{-20}\right) = -20$ . Inputs extremely close to $0$ give outputs far down, but the graph never reaches $x = 0$ .

Increasing and decreasing logarithms

Definition If $a > 1$ , then $\log_a x$ is strictly increasing on $(0, \infty)$ . If $0 < a < 1$ , then $\log_a x$ is strictly decreasing on $(0, \infty)$ . The logarithm has the same direction of change as the exponential with the same base. For $0 < a < 1$ ,

\log_a x = -\log_{1/a} x, \qquad x > 0,

so the graph of $\log_a x$ is the reflection of the graph of $\log_{1/a} x$ across the $x$ -axis.

In plain words A base above $1$ makes both the power and the logarithm go up. A base between $0$ and $1$ makes both go down.

Example Compare base $2$ with base $\frac{1}{2}$ .

$x$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$	$8$
$\log_2 x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$\log_{1/2} x$	$2$	$1$	$0$	$-1$	$-2$	$-3$

Each value in the last row is the negative of the one above it. Both graphs pass through $(1, 0)$ , but the second one falls as $x$ grows and climbs toward $\infty$ as $x \to 0^+$ .

Check yourself

Evaluate $\log_5 125$ , $\log_4 2$ , $\log_{10} 0.001$ and $\log_{1/3} 9$ .

Show the answer

The values are $3$ , $\frac{1}{2}$ , $-3$ and $-2$ , because $5^3 = 125$ , $4^{1/2} = 2$ , $10^{-3} = 0.001$ and $\left(\frac{1}{3}\right)^{-2} = 9$ .

Find the domain and the vertical asymptote of $f(x) = \log_2(6 - 2x)$ .

Show the answer

The argument must be positive, so $6 - 2x > 0$ , which gives $x < 3$ . The domain is $(-\infty, 3)$ . As $x \to 3^-$ the argument approaches $0$ from above, so the line $x = 3$ is the vertical asymptote.

Without a calculator, decide whether $\log_{0.5} 3$ is positive or negative.

Show the answer

It is negative. The base $0.5$ is less than $1$ , so the logarithm is decreasing and equals $0$ at $x = 1$ . Since $3 > 1$ , its value is below $0$ . Indeed $\log_{0.5} 3 = -\log_2 3 \approx -1.585$ .

Logarithms as Inverse Functions

Key terms

Logarithm, base and argument

Exponent as output

Inverse relationship

Positive argument, domain and range

Logarithmic graph

Vertical asymptote

Increasing and decreasing logarithms

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus