Logarithmic Functions

A logarithm answers an exponent question. The statement $\log _b(x)=y$ means that $b^{y}=x$ . For real logarithms, the base must satisfy $b>0$ and $b\ne 1$ , and the input must satisfy $x>0$ .

Logarithmic functions are inverses of exponential functions. That inverse relationship explains both their graphs and the algebraic rules used to solve exponential equations.

On this page (9 sections)

Exponential and logarithmic forms

\log_b x=y\quad\Longleftrightarrow\quad b^y=x

For real logarithms, $b>0$ , $b\ne1$ and $x>0$ . The logarithm is the exponent needed to turn the base into the argument.

\log_2 8=3\quad\text{because}\quad2^3=8

\log_{10}(0.01)=-2\quad\text{because}\quad10^{-2}=0.01

The argument must be positive, but the logarithm’s output can be negative, zero or positive. Every valid base gives $\log_b1=0$ and $\log_bb=1$ .

The common logarithm has base 10. The natural logarithm, written $\ln x$ , has base $e\approx2.71828$ . An unspecified “log” should be interpreted using the stated convention, not guessed.

Logarithms are inverse functions

The graphs of $y=2^x$ and $y=\log_2x$ exchange inputs and outputs. The point $(3,8)$ on the exponential becomes $(8,3)$ on the logarithm.

A logarithm reverses the exponential assignment

$y=2^x$
$y=\log_2x$
Reflection line: $y=x$

Exchanging

(3,8)

with

(8,3)

gives the inverse pair.

Diagram explained

The axes use equal units. The solid exponential and dashed logarithm are reflections across the dotted diagonal. The logarithm has only positive inputs but has negative outputs when 0 < x < 1.

Exact reference values for this graph
Exponential point	Logarithmic point
$(-2,1/4)$	$(1/4,-2)$
$(0,1)$	$(1,0)$
$(1,2)$	$(2,1)$
$(3,8)$	$(8,3)$

The logarithm’s domain is $(0,\infty)$ and its range is all real numbers: any real output $y$ is reached by the positive input $x=b^y$ . The graph passes through $(1,0)$ and has vertical asymptote $x=0$ .

For $b>1$ , the logarithm increases. For 0 < b < 1, it decreases. The inverse relationship remains valid in either case.

Basic logarithm laws

For a valid base $b$ , positive arguments $M,N$ , and real exponent $p$ , the basic laws are:

Logarithm laws with positive arguments
Law	Statement
Product	$\log_b(MN)=\log_bM+\log_bN$
Quotient	$\log_b(M/N)=\log_bM-\log_bN$
Power	$\log_b(M^p)=p\log_bM$

Why the product law works

Write $M=b^r$ and $N=b^s$ . The exponent law gives $MN=b^{r+s}$ , so the logarithm of the product is $r+s$ . Those are precisely the two individual logarithms.

\log_2(8\cdot4)=\log_2 8+\log_2 4=3+2=5

Conditions matter. The identity $\ln(x^2)=2\ln x$ requires $x>0$ . On the larger domain $x\ne0$ , the correct real identity is $\ln(x^2)=2\ln|x|$ .

Solving a logarithmic equation

To solve $\log_2(x-1)=3$ , first require the argument to be positive: $x-1>0$ , so $x>1$ .

\log_2(x-1)=3\quad\Longleftrightarrow\quad x-1=2^3=8

x=9

The input condition holds, and $\log_2(9-1)=\log_2 8=3$ checks the solution. Always test candidates against the arguments in the original logarithmic equation.

Change of base

\log_b x=\frac{\ln x}{\ln b}

If $y=\log_b x$ , then $b^y=x$ . Taking natural logarithms gives $y\ln b=\ln x$ ; division by the non-zero value $\ln b$ yields the formula.

\log_2 8=\frac{\ln8}{\ln2}=3

Any other valid logarithm base works in both numerator and denominator. The base must be the same in both places.

Common logarithm errors

There is no general rule turning a logarithm of a sum into a sum of logarithms. For example, $\log_{10}(10+90)=2$ , whereas $\log_{10}10+\log_{10}90=\log_{10}900$ , which is not 2.

Keep the base, argument and output separate. $\log_2 8$ asks for an exponent; it does not mean $8/2$ . A negative logarithm output is allowed, but zero or negative real arguments are not.

Finally, an exact logarithmic answer need not be replaced by a decimal. Use approximation only when the question or context requires it.

Practice questions

Evaluate $\log _{2}8$ .
Evaluate $\log_{10}(0.01)$ .
Solve $\log _{2}(x-1)=3$ .

Check your answers

$3$ , since $2^3=8$ .
$-2$ , since $10^{-2}=0.01$ .
The argument condition is $x>1$ . Exponential form gives $x-1=8$ , so $x=9$ , which satisfies the condition.

Sources and further reading

OpenStax: Logarithmic functions. Definitions, conventions and related examples.
OpenStax Precalculus 2e. Functions, transformations, inverse functions and function families.
Wolfram Language: FunctionDomain. Domain conventions and symbolic domain checks.
Wolfram Language: FunctionRange. Range conventions and symbolic range checks.