Logarithmic Functions as Inverses, precalculus lesson

Many questions ask for an exponent. Which power of 2 is 1024? At a steady 5 percent growth rate, after how many years has an amount doubled? An exponential function takes an exponent and returns a value. A logarithm runs the machine backwards. It takes the value and returns the exponent.

Key terms

Logarithm as an exponent

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

\begin{gathered} y = \log_b x \\[0.4em] \text{means exactly} \\[0.4em] b^y = x. \end{gathered}

Such a $y$ exists and is unique because the exponential function $y \mapsto b^y$ is one-to-one and takes every positive value.

In plain words The expression $\log_b x$ answers the question “to what power must $b$ be raised to give $x$ ?” A logarithm is an exponent.

Example Each value comes from rewriting the question in exponential form.

Logarithm	Exponential form	Value
$\log_2 32$	$2^5 = 32$	$5$
$\log_3 \frac{1}{9}$	$3^{-2} = \frac{1}{9}$	$-2$
$\log_5 1$	$5^0 = 1$	$0$
$\log_9 3$	$9^{1/2} = 3$	$\frac{1}{2}$
$\log_{1/2} 8$	$\left(\frac{1}{2}\right)^{-3} = 8$	$-3$

Positive base other than one

Definition The base of a logarithm must satisfy $b > 0$ and $b \ne 1$ . These are exactly the bases for which $b^y$ is a real number for every real $y$ and the function $y \mapsto b^y$ is one-to-one onto $(0, \infty)$ . The other bases fail for different reasons.

For the base $1$ , the power $1^y$ equals $1$ for every $y$ , so the equation $1^y = 1$ has every real number as a solution and $1^y = 5$ has none.
For the base $0$ , the power $0^y$ is $0$ when $y > 0$ and is undefined when $y \le 0$ .
For a negative base, the power is not a real number for many exponents, for example $(-4)^{1/2}$ .

In plain words The base must be a positive number other than 1. Any other choice either gives no answer, too many answers, or answers that are not real numbers.

Example Is $\log_{-2} 4 = 2$ , since $(-2)^2 = 4$ ? No. The function $y \mapsto (-2)^y$ is undefined at $y = \frac{1}{2}$ , because $-2$ has no real square root, and at many other inputs, so negative bases are left out entirely and $\log_{-2} 4$ is undefined.

Positive argument

Definition The number $x$ in $\log_b x$ is called the argument. For a valid base $b$ , the power $b^y$ is positive for every real $y$ , so the equation $b^y = x$ has a solution only when $x > 0$ . Therefore $\log_b x$ is defined exactly for $x > 0$ . The logarithm of $0$ and the logarithm of a negative number are undefined in the real numbers.

In plain words The input of a logarithm must be positive. The output can be any real number, including zero and negative numbers.

Example The expression $\log_{10}(-100)$ is undefined, because $10^y$ is never negative, and $\log_2 0$ is undefined, because $2^y$ is never $0$ . On the other hand $\log_2 0.25 = -2$ is fine, a negative output from a positive input. For the expression $\log_3(x - 4)$ to make sense, the argument must be positive, so $x - 4 > 0$ and $x > 4$ .

Inverse relationship

Definition For $b > 0$ with $b \ne 1$ , the function $f(x) = b^x$ is one-to-one with domain $\mathbb{R}$ and range $(0, \infty)$ . Its inverse is the logarithmic function $f^{-1}(x) = \log_b x$ . The two functions undo each other,

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

A point $(a, c)$ lies on the graph of $y = b^x$ exactly when $(c, a)$ lies on the graph of $y = \log_b x$ , so the two graphs are mirror images in the line $y = x$ .

In plain words Raising $b$ to a power and taking $\log_b$ cancel each other, as long as every logarithm you write has a positive argument.

Example By the first identity $\log_7\left(7^{2.5}\right) = 2.5$ , and by the second $10^{\log_{10} 42} = 42$ . The points $\left(-1, \frac{1}{3}\right)$ , $(0, 1)$ and $(2, 9)$ lie on $y = 3^x$ , so the points $\left(\frac{1}{3}, -1\right)$ , $(1, 0)$ and $(9, 2)$ lie on $y = \log_3 x$ . The second identity does not apply to $x = -5$ , since $\log_b(-5)$ is undefined.

Common logarithm

Definition The common logarithm is the logarithm with base $10$ , written without a base,

\log x = \log_{10} x, \qquad x > 0.

So $\log x = y$ means $10^y = x$ , and $\log\left(10^k\right) = k$ for every real $k$ . Calculators use this convention. Some advanced texts and software write $\log$ for the natural logarithm instead, so check the convention when you switch sources.

In plain words The common logarithm counts powers of ten. A positive whole number with $d$ digits has a common logarithm that is at least $d - 1$ and less than $d$ .

Example From the powers of ten, $\log 1000 = 3$ and $\log 0.001 = -3$ . Since $10 < 50 < 100$ , the value $\log 50$ lies between $1$ and $2$ , and a calculator gives $\log 50 \approx 1.69897$ . The number $3600000$ has $7$ digits, and indeed $\log 3600000 \approx 6.556$ lies between $6$ and $7$ .

Natural logarithm

Definition The natural logarithm is the logarithm with base $e \approx 2.71828$ ,

\ln x = \log_e x, \qquad x > 0.

It is the inverse of the natural exponential function, so $\ln x = y$ means $e^y = x$ . In particular $\ln 1 = 0$ and $\ln e = 1$ , and

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

In plain words The value $\ln x$ is the exponent you put on $e$ to get $x$ . It is the logarithm that shows up in continuous growth and throughout calculus.

Example An amount grows continuously at 5 percent per year, so it is multiplied by $e^{0.05t}$ after $t$ years. It has doubled when $e^{0.05t} = 2$ . By the definition of $\ln$ this means $0.05t = \ln 2$ , so

t = \frac{\ln 2}{0.05} \approx 13.86 \text{ years.}

Domain and range

Definition For $b > 0$ with $b \ne 1$ , the function $\log_b x$ has domain $(0, \infty)$ and range $\mathbb{R}$ , the set of all real numbers. These are the domain and range of $b^x$ with the roles swapped, as for every inverse function. When the argument is an expression $g(x)$ , the natural domain of $\log_b g(x)$ is the set of all $x$ in the domain of $g$ with $g(x) > 0$ .

In plain words Inputs must be positive, and outputs can be anything. Every real number $y$ is a logarithm of something, namely $y = \log_b\left(b^y\right)$ .

Example For $f(x) = \log_2(6 - 2x)$ the argument must be positive, so $6 - 2x > 0$ , which gives $x < 3$ . The domain is $(-\infty, 3)$ . To see that the range is all of $\mathbb{R}$ , pick any output $y$ and solve $6 - 2x = 2^y$ , which gives

x = \frac{6 - 2^y}{2},

a number below $3$ . For instance the output $5$ comes from $x = -13$ , and the output $-1$ comes from $x = \frac{11}{4}$ .

Check yourself

Evaluate $\log_4 64$ , $\log 0.01$ , $\ln\left(e^{-3}\right)$ and $\log_8 2$ .

Show the answer

The values are $3$ , $-2$ , $-3$ and $\frac{1}{3}$ , because $4^3 = 64$ , $10^{-2} = 0.01$ , $\ln\left(e^x\right) = x$ for every real $x$ , and $8^{1/3} = 2$ .

Find the domain of $f(x) = \ln(5 - x) + \log(x + 2)$ .

Show the answer

Both arguments must be positive. The first needs $x < 5$ and the second needs $x > -2$ , so the domain is the interval $(-2, 5)$ .

Solve $e^{2t} = 7$ exactly, then give a decimal value.

Show the answer

By the definition of the natural logarithm $2t = \ln 7$ , so

t = \frac{\ln 7}{2} \approx 0.973.

Logarithmic Functions as Inverses

Key terms

Logarithm as an exponent

Positive base other than one

Positive argument

Inverse relationship

Common logarithm

Natural logarithm

Domain and range

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus