Exponentials and Logarithms

Distinguish additive change from exponential growth and decay, then read logarithms as inverse exponents. Use domains and logarithm rules carefully when moving between products, sums, powers, and differences.

An exponential function places the changing quantity in the exponent, so equal steps in the input produce equal multiplying factors in the output. A logarithm reverses that relationship by asking which exponent produced a given positive value.

Exponential Change Uses a Fixed Factor

Compare the linear function

f(x)=2x

with the exponential function

g(x)=2^x.

Increasing $x$ by $1$ adds $2$ to $f(x)$ , but it multiplies $g(x)$ by $2$ :

$x$	$2x$	$2^x$
0	0	1
1	2	2
2	4	4
3	6	8
4	8	16

Linear change adds a fixed amount. Exponential change multiplies by a fixed factor. More generally, if $g(x)=b^x$ , then:

g(x+1)=b^{x+1}=b\cdot b^x=b\,g(x).

For a real-valued exponential function defined for every real input, the base must satisfy $b>0$ . The behavior depends on the base:

if $b>1$ , the function grows as $x$ increases;
if $0<b<1$ , the function decays as $x$ increases;
if $b=1$ , the function is constant because $1^x=1$ .

A negative base can be raised to integer powers, but it does not produce a real value for every real exponent. It therefore does not define the same kind of real exponential function on all real inputs.

Exercise: Continue exponential growth

If $g(x)=3^x$ , what is $g(4)$ ?

Your answer

Compute it first, then check your number.

HintExpand the power

Write four factors of $3$ .

SolutionMultiply by the fixed factor

Evaluate the power:

3^4=3\cdot3\cdot3\cdot3=81.

Each increase of $1$ in the exponent multiplies the output by $3$ .

Exercise: Trace exponential decay

If $h(x)=\left(\frac{1}{2}\right)^x$ , what is $h(3)$ ?

Your answer

Compute it first, then check your number.

HintUse the base as the repeated factor

Expand the power as $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}$ .

SolutionMultiply by one-half at each step

The value is:

h(3)=\left(\frac{1}{2}\right)^3 =\frac{1}{8} =0.125.

Because the base lies between $0$ and $1$ , each increase in $x$ makes the output smaller.

Why Base (e) Appears Often

The constant

e\approx2.71828

is a particularly useful exponential base. The function $e^x$ , also written $\exp(x)$ , describes continuous proportional change particularly cleanly. Later, calculus will show that its rate of change at every point is equal to its current value.

For every real $x$ , $e^x$ is positive, $e^0=1$ , and:

e^{-x}=\frac{1}{e^x}.

These properties make the natural exponential useful in growth and decay, probability, differential equations, and optimization. In machine learning, it can also turn unrestricted real scores into positive quantities.

Logarithms Undo Exponentials

For a base $b>0$ with $b\ne1$ , the statement

\log_b(y)=x

means exactly the same thing as:

b^x=y.

Log base $b$ of $y$ equals $x$ exactly when $b$ raised to the power $x$ equals $y$ .

The input must satisfy $y>0$ because a positive base raised to a real power always produces a positive value.

For example:

\log_2(8)=3

because $2^3=8$ .

Logarithms can be zero or negative even though their inputs must be positive:

\log_{10}(1)=0,\qquad \log_{10}(0.01)=-2.

The natural logarithm, written $\ln(y)$ , uses base $e$ . Some mathematical texts also use $\log(y)$ for the natural logarithm, while calculators and other fields may use $\log(y)$ for base $10$ . A formula should state its convention when the base is not clear.

Exercise: Read a logarithm as an exponent

What is $\log_{10}(1000)$ ?

Your answer

Compute it first, then check your number.

HintRewrite as a power

Find $x$ such that $10^x=1000$ .

SolutionFind the required exponent

Rewrite the logarithmic statement as an exponential one:

10^x=1000.

Since $10^3=1000$ , the required exponent is $3$ . Therefore:

\log_{10}(1000)=3.

Logarithms Turn Products into Sums

For positive $a$ and $c$ , logarithms with the same valid base satisfy:

\log_b(ac)=\log_b(a)+\log_b(c).

Repeated products therefore become sums. Using the natural logarithm:

\ln\left(\prod_i p_i\right)=\sum_i\ln(p_i).

This form is useful whenever many positive factors are multiplied. For example, the direct product of many small probabilities may become too small for a computer to represent accurately, while the sum of their logarithms remains manageable.

Division becomes subtraction:

\log_b\left(\frac{a}{c}\right)=\log_b(a)-\log_b(c),

and powers become multiplication:

\log_b(a^k)=k\log_b(a).

These rules require positive logarithm inputs. They also do not turn addition into a simpler operation: in general, $\log_b(a+c)\ne\log_b(a)+\log_b(c)$ .

Exercise: Convert a product to a log sum

Which expression equals $\log(p_1p_2p_3)$ for positive probabilities?

Choose the equivalent expression

log(p₁) + log(p₂) + log(p₃)

log(p₁) · log(p₂) · log(p₃)

(log(p₁) + log(p₂) + log(p₃)) / 3

Select one choice, then check.

HintUse the product rule repeatedly

First separate $p_1$ from $p_2p_3$ , then separate the remaining product.

SolutionApply the logarithm product rule

Repeated use of the product rule gives:

\log(p_1p_2p_3) =\log(p_1)+\log(p_2)+\log(p_3).

Every probability is positive, and the same logarithm base is used in each term. No averaging occurs unless a later formula explicitly divides by the number of terms.

Exercise: Check a logarithm input

Which expression is defined as a real logarithm?

Choose the valid input

ln(0)

ln(-2)

ln(0.2)

Select one choice, then check.

HintUse the range of an exponential

The value $e^x$ is positive for every real $x$ . A logarithm can therefore reverse it only for a positive input.

SolutionRequire a positive input

The expression $\ln(0.2)$ is defined because $0.2>0$ . Its value is negative because $0.2$ lies between $0$ and $1$ .

Neither $\ln(0)$ nor $\ln(-2)$ has a real value.

Use the Inverse Relationship as a Check

For positive $y$ :

\exp(\ln y)=y,

and for every real $x$ :

\ln(\exp x)=x.

The conditions differ because $\ln y$ requires $y>0$ , while $\exp x$ accepts every real $x$ . When exponentials and logarithms appear together, identify the base and domain before using the inverse relationship.

An exponential asks for the value produced by an exponent. A logarithm asks for the exponent that produced a positive value. Moving between these two forms is often the simplest way to evaluate an expression, solve an equation, or check a result.