Laws of Logarithms

Logarithms are exponents, so they follow rules that come straight from the exponent laws. These laws of logarithms let you combine several logs into one, split one log into simpler pieces, evaluate a log in any base on a calculator, and (on the next pages) solve equations.

Key ideas

Four basic facts

For any base $b \gt 0$ , $b \ne 1$ :

Fact	Why
$\log_b 1 = 0$	$b^0 = 1$
$\log_b b = 1$	$b^1 = b$
$\log_b\left(b^x\right) = x$	the exponent on $b$ is $x$
$b^{\log_b x} = x$ (for $x \gt 0$ )	$\log_b x$ is the exponent that turns $b$ into $x$

The product law

\log_b(xy) = \log_b x + \log_b y \qquad (x, y \gt 0)

Why it works. Let $m = \log_b x$ and $n = \log_b y$ , so $x = b^m$ and $y = b^n$ . When you multiply powers, you add exponents:

xy = b^m \cdot b^n = b^{m + n} \quad\Rightarrow\quad \log_b(xy) = m + n = \log_b x + \log_b y

For example, $\log 1000 = \log(10 \times 100) = \log 10 + \log 100 = 1 + 2 = 3$ . ✓

The quotient law

\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y \qquad (x, y \gt 0)

This comes from dividing powers: $\dfrac{b^m}{b^n} = b^{m - n}$ . For example, $\log_2 32 - \log_2 4 = 5 - 2 = 3$ , and $\log_2 \dfrac{32}{4} = \log_2 8 = 3$ . ✓

The power law

\log_b\left(x^n\right) = n\log_b x \qquad (x \gt 0)

This comes from a power of a power: if $x = b^m$ , then $x^n = \left(b^m\right)^n = b^{mn}$ , so $\log_b\left(x^n\right) = mn = n\log_b x$ . The power law also handles roots, since $\sqrt{x} = x^{\frac{1}{2}}$ .

Exponent law	Matching log law
$b^m \cdot b^n = b^{m + n}$	$\log_b(xy) = \log_b x + \log_b y$
$\dfrac{b^m}{b^n} = b^{m - n}$	$\log_b \dfrac{x}{y} = \log_b x - \log_b y$
$\left(b^m\right)^n = b^{mn}$	$\log_b\left(x^n\right) = n\log_b x$

Change of base

Your calculator has a LOG key (base $10$ ), but not a key for every base. To evaluate a log in any base:

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

Why it works. Let $y = \log_b x$ , so $b^y = x$ . Take the common log of both sides and use the power law:

\log\left(b^y\right) = \log x \quad\Rightarrow\quad y\log b = \log x \quad\Rightarrow\quad y = \frac{\log x}{\log b}

This replaces the systematic trial from the introduction to logarithms.

What the laws do not say

There’s no law for the log of a sum or difference, and dividing two logs is not the quotient law:

\log(x + y) \ne \log x + \log y \qquad\qquad \frac{\log x}{\log y} \ne \log\left(\frac{x}{y}\right)

Also, the laws only combine logs with the same base.

Worked examples

Example 1: Evaluating with the product and quotient laws

Evaluate without a calculator.

(a) $\log_6 4 + \log_6 9$
(b) $\log_2 48 - \log_2 3$
(c) $3\log_5 10 - \log_5 8$

Solution. Neither log in each part is a nice number on its own, so combine first.

(a) $\log_6 4 + \log_6 9 = \log_6(4 \times 9) = \log_6 36 = 2$

(b) $\log_2 48 - \log_2 3 = \log_2 \dfrac{48}{3} = \log_2 16 = 4$

3\log_5 10 - \log_5 8 = \log_5\left(10^3\right) - \log_5 8 = \log_5 \frac{1000}{8} = \log_5 125 = 3

Example 2: Powers and roots

Evaluate.

(a) $\log_3 \sqrt{27}$
(b) $\log_2\left(8^5\right)$
(c) $\log_5 \sqrt[3]{25}$

Solution.

(a) $\log_3 \sqrt{27} = \log_3\left(27^{\frac{1}{2}}\right) = \tfrac{1}{2}\log_3 27 = \tfrac{1}{2}(3) = \tfrac{3}{2}$

(b) $\log_2\left(8^5\right) = 5\log_2 8 = 5(3) = 15$

Example 3: Change of base

Evaluate $\log_7 200$ to two decimal places, and check your answer.

Solution.

\log_7 200 = \frac{\log 200}{\log 7} \approx \frac{2.3010}{0.8451} \approx 2.72

It’s reasonable: $7^2 = 49$ and $7^3 = 343$ , so the answer should be between $2$ and $3$ , closer to $3$ .

Check: $7^{2.72} \approx 198.9 \approx 200$ . ✓

Example 4: Single logs and expanded logs

(a) Write $2\log x + \log y - 3\log z$ as a single logarithm.
(b) Expand $\log_2 \dfrac{8x^3}{\sqrt{y}}$ as fully as possible.
(c) Simplify $\log_3\left(9^x \cdot 3\right)$ .

Assume $x$ , $y$ , $z \gt 0$ .

Solution.

(a) Move the coefficients up as exponents (power law), then combine (product and quotient laws):

2\log x + \log y - 3\log z = \log\left(x^2\right) + \log y - \log\left(z^3\right) = \log \frac{x^2 y}{z^3}

(b) Split the quotient and product, then use the power law:

\begin{aligned} \log_2 \frac{8x^3}{\sqrt{y}} &= \log_2 8 + \log_2\left(x^3\right) - \log_2\left(y^{\frac{1}{2}}\right) \\ &= 3 + 3\log_2 x - \tfrac{1}{2}\log_2 y \end{aligned}

(c) Write everything as a power of $3$ : $9^x \cdot 3 = 3^{2x} \cdot 3^1 = 3^{2x + 1}$ . So $\log_3\left(9^x \cdot 3\right) = 2x + 1$ .

Common mistakes

Splitting the log of a sum. $\log(x + y)$ can’t be simplified. Test it: $\log(10 + 10) = \log 20 \approx 1.30$ , but $\log 10 + \log 10 = 2$ . The product law is about $\log(xy)$ , a product inside the log.

Treating a quotient of logs like the quotient law. $\dfrac{\log 100}{\log 10} = \dfrac{2}{1} = 2$ , but $\log \dfrac{100}{10} = \log 10 = 1$ . A fraction of two logs is what the change of base formula gives you, not a single log.

Applying the power law to the wrong thing. The power law needs the whole argument raised to the power. $\log\left(3x^2\right) = \log 3 + 2\log x$ , but $\log\left((3x)^2\right) = 2\log(3x)$ . And $(\log x)^2$ is not $2\log x$ .

Combining logs with different bases. $\log_2 8 + \log_3 9$ is just $3 + 2 = 5$ . You can’t write it as $\log_6 72$ or anything like that.

Turning change of base upside down. It’s $\log_b x = \dfrac{\log x}{\log b}$ : the number goes on top, the base on the bottom. A quick sense check (like “is $\log_7 200$ between $2$ and $3$ ?”) catches this.

Mixing up $\log_b 1$ and $\log_b b$ . $\log_b 1 = 0$ (because $b^0 = 1$ ) and $\log_b b = 1$ (because $b^1 = b$ ).

Practice

1. (Warm-up) Evaluate $\log_4 2 + \log_4 8$ .

Solution

$\log_4 2 + \log_4 8 = \log_4 16 = 2$

2. (Warm-up) Evaluate $\log_3 162 - \log_3 2$ .

Solution

$\log_3 162 - \log_3 2 = \log_3 \dfrac{162}{2} = \log_3 81 = 4$

3. (Warm-up) Evaluate.

(a) $\log_7 7$
(b) $\log_9 1$
(c) $5^{\log_5 11}$

Solution

(a) $1$

(b) $0$

4. (Core) Evaluate without a calculator.

(a) $2\log 5 + \log 4$
(b) $\log_2 40 - \log_2 5 + \log_2 4$

Solution

(a) $2\log 5 + \log 4 = \log 25 + \log 4 = \log 100 = 2$

(b) $\log_2 \dfrac{40 \times 4}{5} = \log_2 32 = 5$

5. (Core) Write $3\log_2 x + \log_2 y - \tfrac{1}{2}\log_2 z$ as a single logarithm. Assume $x$ , $y$ , $z \gt 0$ .

Solution

\log_2\left(x^3\right) + \log_2 y - \log_2\left(z^{\frac{1}{2}}\right) = \log_2 \frac{x^3 y}{\sqrt{z}}

6. (Core) Expand $\log \dfrac{100x^4}{y^2}$ fully. Assume $x$ , $y \gt 0$ .

Solution

\log 100 + \log\left(x^4\right) - \log\left(y^2\right) = 2 + 4\log x - 2\log y

7. (Core) Use the change of base formula to evaluate each to three decimal places.

(a) $\log_3 50$
(b) $\log_{0.5} 10$

Solution

(a) $\log_3 50 = \dfrac{\log 50}{\log 3} \approx 3.561$ . Check: $3^3 = 27$ and $3^4 = 81$ , so it should be between $3$ and $4$ . ✓

(b) $\log_{0.5} 10 = \dfrac{\log 10}{\log 0.5} = \dfrac{1}{\log 0.5} \approx -3.322$ . It’s negative because the base is less than $1$ and $10 \gt 1$ . Check: $0.5^{-3.322} \approx 10$ . ✓

8. (Challenge) Suppose $\log_b 2 \approx 0.356$ and $\log_b 3 \approx 0.565$ . Use the laws of logarithms to estimate:

(a) $\log_b 12$
(b) $\log_b 4.5$
(c) $\log_b \sqrt{6}$

Solution

(a) $12 = 2^2 \times 3$ , so $\log_b 12 = 2\log_b 2 + \log_b 3 \approx 2(0.356) + 0.565 = 1.277$ .

(b) $4.5 = \dfrac{9}{2} = \dfrac{3^2}{2}$ , so $\log_b 4.5 = 2\log_b 3 - \log_b 2 \approx 1.130 - 0.356 = 0.774$ .

(c) $\log_b \sqrt{6} = \tfrac{1}{2}\log_b 6 = \tfrac{1}{2}\left(\log_b 2 + \log_b 3\right) \approx \tfrac{1}{2}(0.921) \approx 0.461$ .

(These values fit $b = 7$ : $\log_7 12 \approx 1.277$ .)

9. (Challenge) Simplify.

(a) $\log_3 \dfrac{27^x}{9}$
(b) $10^{2\log 5}$
(c) Show that $\log_4 x = \tfrac{1}{2}\log_2 x$ for $x \gt 0$ .

Solution

(a) $\dfrac{27^x}{9} = \dfrac{3^{3x}}{3^2} = 3^{3x - 2}$ , so $\log_3 \dfrac{27^x}{9} = 3x - 2$ .

(b) $2\log 5 = \log\left(5^2\right) = \log 25$ , so $10^{2\log 5} = 10^{\log 25} = 25$ .

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} = \tfrac{1}{2}\log_2 x