Math Core

Logarithms are exponents, so every rule for exponents has a matching rule for logarithms. These properties are what make logs useful: they turn products into sums and powers into multiples, and they let you rewrite a log in any base using only $\ln$ or $\log$ .

From exponent rules to log rules

Let $M = b^m$ and $N = b^n$ , so that $\log_b M = m$ and $\log_b N = n$ . The exponent rules give

MN = b^{m+n}, \qquad \frac{M}{N} = b^{m-n}, \qquad M^p = b^{mp}.

Now read each equation as a logarithm statement. The first says $\log_b(MN) = m + n$ , the second says $\log_b\dfrac{M}{N} = m - n$ , and the third says $\log_b\left(M^p\right) = pm$ . Substituting back $m = \log_b M$ and $n = \log_b N$ gives the three properties.

Properties of logarithms

For $b > 0$ , $b \ne 1$ , positive numbers $M$ and $N$ , and any real $p$ :

Property	Rule	In words
Product	$\log_b(MN) = \log_b M + \log_b N$	log of a product is a sum
Quotient	$\log_b\dfrac{M}{N} = \log_b M - \log_b N$	log of a quotient is a difference
Power	$\log_b\left(M^p\right) = p\log_b M$	an exponent comes out as a factor

A quick numerical check: $\log_2 4 + \log_2 8 = 2 + 3 = 5$ , and $\log_2(4 \cdot 8) = \log_2 32 = 5$ . ✓

Common mistake

The properties say nothing about sums inside a log. $\ln(x + y)$ is not $\ln x + \ln y$ , and it cannot be simplified. Also, $\dfrac{\log_b M}{\log_b N}$ is not $\log_b\dfrac{M}{N}$ and $(\log_b M)^2$ is not $2\log_b M$ . Only products, quotients and powers inside the logarithm can be broken apart.

Expanding

To expand a logarithm, rewrite it so that no log contains a product, quotient, power or root. Convert roots to fractional exponents first ( $\sqrt[3]{x} = x^{1/3}$ ), then work from the outside in: quotient, then product, then power.

Worked example: Expand a logarithm

Expand $\log_2 \dfrac{8x^3}{\sqrt{y}}$ . Assume $x, y > 0$ .

Solution.

\begin{aligned} \log_2 \frac{8x^3}{y^{1/2}} &= \log_2\left(8x^3\right) - \log_2 y^{1/2} && \text{quotient} \\ &= \log_2 8 + \log_2 x^3 - \log_2 y^{1/2} && \text{product} \\ &= 3 + 3\log_2 x - \tfrac{1}{2}\log_2 y && \text{power, and } \log_2 8 = 3 \end{aligned}

Expanding is useful when a messy log appears in a formula; in calculus it makes certain derivatives much easier. It also lets you compute new logs from a few known ones.

Worked example: Build logs from known values

Suppose $\log_b 2 = 0.39$ and $\log_b 5 = 0.90$ . Find $\log_b 20$ and $\log_b 0.4$ .

Solution. Write each number with $2$ 's and $5$ 's. Since $20 = 2^2 \cdot 5$ ,

\log_b 20 = 2\log_b 2 + \log_b 5 = 2(0.39) + 0.90 = 1.68.

Since $0.4 = \dfrac{2}{5}$ ,

\log_b 0.4 = \log_b 2 - \log_b 5 = 0.39 - 0.90 = -0.51.

The negative answer makes sense: $0.4 < 1$ , and the log of a number less than $1$ is negative when $b > 1$ .

Condensing

To condense, run the properties in reverse and write the expression as a single logarithm. The order matters: move every coefficient up as an exponent first, then combine. Added terms go in the numerator, subtracted terms in the denominator.

Worked example: Condense to a single log

Write $2\ln x - \tfrac{1}{3}\ln(x + 4) + \ln 5$ as a single logarithm.

Solution.

\begin{aligned} 2\ln x - \tfrac{1}{3}\ln(x + 4) + \ln 5 &= \ln x^2 - \ln(x + 4)^{1/3} + \ln 5 && \text{power} \\ &= \ln\left(5x^2\right) - \ln\sqrt[3]{x + 4} && \text{product} \\ &= \ln\frac{5x^2}{\sqrt[3]{x + 4}} && \text{quotient} \end{aligned}

A subtlety about domain

The properties require positive inputs. That matters when variables are involved. For example, $\ln\left(x^2\right)$ is defined for every $x \ne 0$ , but $2\ln x$ is defined only for $x > 0$ . The two expressions agree when $x > 0$ and differ otherwise; the correct identity for all $x \ne 0$ is $\ln\left(x^2\right) = 2\ln|x|$ . When you expand or condense, keep track of where the original expression was defined, especially when solving equations in the next lesson.

Change of base

Your calculator has only $\log$ and $\ln$ . To evaluate a log in any other base, convert it.

Change-of-base formula

For any positive $x$ and bases $a, b$ (positive, not $1$ ):

\log_b x = \frac{\log_a x}{\log_a b}, \qquad\text{in particular}\qquad \log_b x = \frac{\ln x}{\ln b} = \frac{\log x}{\log b}.

Why it works. Let $y = \log_b x$ , so $b^y = x$ . Take $\ln$ of both sides and apply the power property: $y\ln b = \ln x$ . Divide by $\ln b$ .

Worked example: Change of base

(a) Evaluate $\log_3 20$ to three decimal places.

(b) Find the exact value of $\log_4 32$ .

Solution.

(a) $\log_3 20 = \dfrac{\ln 20}{\ln 3} \approx \dfrac{2.99573}{1.09861} \approx 2.727$ . This is reasonable because $3^2 = 9$ and $3^3 = 27$ , so the answer lies between $2$ and $3$ .

(b) Use base $2$ , since $4$ and $32$ are both powers of $2$ :

\log_4 32 = \frac{\log_2 32}{\log_2 4} = \frac{5}{2}.

Check: $4^{5/2} = \left(\sqrt{4}\right)^5 = 32$ . ✓

Change of base also shows that every logarithmic function is a vertical stretch of $\ln x$ : $\log_3 x = \dfrac{1}{\ln 3}\ln x \approx 0.910\ln x$ . That's how you graph $y = \log_3 x$ on a grapher with no base- $3$ key.

y = log₃ x = (ln x)/(ln 3) is a vertical compression of y = ln x.Open in grapher →

Rewriting exponentials in base e

The same idea converts any exponential to base $e$ . Since $b = e^{\ln b}$ ,

b^x = \left(e^{\ln b}\right)^x = e^{(\ln b)x}.

For example, $2^x = e^{0.6931x}$ and $(0.8)^x = e^{-0.2231x}$ . A base greater than $1$ gives a positive coefficient $k = \ln b$ ; a base between $0$ and $1$ gives a negative one. You'll use this in the modeling lesson, where growth is often written as $e^{kt}$ .

Tip

Estimate before you compute. If $\log_7 90$ comes out as $0.43$ on your calculator, something is wrong: $7^2 = 49$ and $7^3 = 343$ , so the answer must be between $2$ and $3$ . You probably divided in the wrong order.

Practice

Practice 1

Evaluate $\log_{12} 3 + \log_{12} 48$ without a calculator.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Evaluate $\log_5 250 - \log_5 2$ without a calculator.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which is the expansion of $\ln\dfrac{x^2\sqrt{x + 1}}{y^3}$ ? Assume all expressions are defined.

Practice 4

Which single logarithm equals $3\log x + \log 4 - \dfrac{1}{2}\log y$ ?

Practice 5

If $\log_b x = 4$ , $\log_b y = -1$ and $\log_b z = 2$ , find $\log_b\dfrac{x^3}{yz^2}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Use the change-of-base formula to evaluate $\log_7 90$ . Round to three decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Find the exact value of $\log_8 32$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Evaluate $\log_2 3 \cdot \log_3 16$ without a calculator.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.