Laws of logarithms

In a nutshell

Similar to the laws for exponentials, e.g. $a^x \times a^y = a^{x + y}$ , laws for logarithms exist which you can use to manipulate expressions they feature in. Here you will derive and cover those laws.

Multiplication, division and power laws

Multiplication law

Let $a$ be a positive constant not equal to $1$ , and let $m$ and $n$ be any real values. Suppose that $a^m = x$ , and $a^n = y$ . Then the laws of exponentials tell you that:

$\begin{aligned}xy &= a^m \times a^n\\&= a^{m+n}\end{aligned}$

Applying $\log_a()$ to $x$ , $y$ , and $xy$ , you have that:

$\begin{aligned}m &= \log_a(x)\\n &= \log_a(y)\\m+n &= \log_a(xy)\end{aligned}$

Put these together to derive the multiplication law for logarithms:

$\boxed{\log_a(xy) = \log_a(x) + \log_a(y)}$

Division law

If $x$ and $y$ are constants with $y \neq 0$ , then you have that $\dfrac{x}{y} \times y = x$ . Apply $\log_a()$ to this equation, use the multiplication law:

$\begin{aligned}\log_a\left(\dfrac{x}{y} \times y \right) &= \log_a(x)\\\log_a\left(\dfrac{x}{y}\right) + \log_a(y) &= \log_a(x)\end{aligned}$

Rearrange this to derive the division law for logarithms:

$\boxed{\log_a\left(\dfrac{x}{y}\right)= \log_a(x) - \log_a(y)}$

Power law

You have that $x = a^m$ . Laws of exponentials tell you that:

$\begin{aligned}x^k &= (a^m)^k\\&= a^{km}\end{aligned}$

Apply $\log_a()$ to get:

$\log_a(x^k) = km$

Substitute $m = \log_a(x)$ to derive the power law for logarithms:

$\boxed{\log_a(x^k) = k\log_a(x)}$

Example 1

Let $a$ be a positive constant not equal to $1$ , and let $x \ge 0$ . Show that $\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}$ .

You have that $\sqrt{x} = x^{\frac12}$ . The power law for logarithms tells you that:

$\begin{aligned}\sqrt{x} &= x^{\frac12}\\\log_a(\sqrt{x}) &= \log_a(x^\frac12)\\\log_a(\sqrt{x}) &= \dfrac12\log_a(x)\end{aligned}$

Therefore, $\underline{\log_a(\sqrt{x}) = \dfrac{\log_a(x)}{2}}$ .

Special cases

For a positive constant $a$ not equal to $1$ , recall that the unique value of $x$ such that $a^x = 1$ is $x = 0$ . Similarly, the unique value of $x$ for which $a^x = a$ is $x = 1$ . Therefore, you have the following special cases:

$\begin{aligned}\log_a(1) &= 0\\\log_a(a) &= 1\end{aligned}$

Using this and the division law, you can obtain the following useful case:

$\begin{aligned}\log_a\left(\dfrac1x\right) & = \log_a(1) - \log_a(x)\\&= -\log_a(x)\end{aligned}$

Example 2

WIthout using a calculator, find the exact value of $\log_3\left(\dfrac{\sqrt{27}}{9}\right)$ .

You have that $\sqrt{27} = 3^{\frac32}$ , and $9 = 3^2$ . Use the power and division laws to get:

$\begin{aligned}\log_3\left(\dfrac{\sqrt{27}}{9}\right)&= \log_3(\sqrt{27}) - \log_3(9)\\&= \log_3(3^{\frac32}) - \log_3(3^2)\\&= \dfrac32\log_3(3) - 2\log_3(3)\\&= \dfrac{3}{2} - 2\\&= -\dfrac12\end{aligned}$

Therefore, the exact value of $\log_3\left(\dfrac{\sqrt{27}}{9}\right)$ is $\underline{-\dfrac12}$ .