Laws of Logarithms

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Laws of Logarithmsvideo

Laws of Logarithms

Laws of logarithms

What are the laws of logarithms?

Laws of logarithms allow you to simplify and manipulate expressions involving logarithms
- The laws of logarithms are equivalent to the laws of indices
The laws you need to know are, given $a, x, y > 0$ :
- $\log_{a} x y = \log_{a} x + \log_{a} y$
  - This relates to $a^{x} \times a^{y} = a^{x + y}$
- $\log_{a} \frac{x}{y} = \log_{a} x {- \log}_{a} y$
  - This relates to $a^{x} \div a^{y} = a^{x - y}$
- $\log_{a} x^{m} = m \log_{a} x$
  - This relates to $(a^{x})^{y} = a^{x y}$

There are also some particular results these lead to
- $\log_{a} a = 1$
- $\log_{a} a^{x} = x$
- $a^{\log_{a} x} = x$
- $\log_{a} 1 = 0$
- $\log_{a} (\frac{1}{x}) = - \log_{a} x$

Beware…
- … $\log_{a} (x + y) \neq \log_{a} x + \log_{a} y$
These results apply to $\ln x (\log_{e} x)$ too
- Two particularly useful results are
  - $\ln e^{x} = x$
  - $e^{\ln x} = x$

How do I use the laws of logarithms?

Laws of logarithms can be used to …
- … simplify expressions
- … solve logarithmic equations
- … solve exponential equations

Simplifying an expression using the laws of logarithms

Worked example

a) Write the expression $2 \log 4 - \log 2$ in the form $\log k$ , where $k \in ℤ$ .

Using the law $\log_{a} x^{m} = m \log_{a} x$ we can rewrite $2 \log 4$ as $\log 4^{2} = \log 16$ .

$2 \log 4 - \log 2 = \log 16 - \log 2$

Using the law $\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$ :

$\log 16 - \log 2 = \log \frac{16}{2} = \log 8$

$2 \log 4 - \log 2 = \log 8$

b) Hence, or otherwise, solve $2 \log 4 - \log 2 = - \log \frac{1}{x}$ .

Rewrite the equation using the expression found in part (a).

$2 \log 4 - \log 2 = \log 8$ :

$\log 8 = - \log \frac{1}{x}$

Using the index law $\frac{1}{x} = x^{- 1}$ ::

$\log 8 = - \log x^{- 1}$

Using the law $\log_{a} x^{m} = m \log_{a} x$ :

$\log 8 = \log x^{- (- 1)} = \log x$

Compare the two sides.

$\log 8 = \log x$

$x = 8$

Change of Base

Change of base

How do I change the base of a logarithm?

The formula for changing the base of a logarithm is

$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$

The value you choose for b does not matter, however if you do not have a calculator, you can choose b such that the problem will be possible to solve

Why change the base of a logarithm?

The laws of logarithms can only be used if the logs have the same base
- If a problem involves logarithms with different bases, you can change the base of the logarithm and then apply the laws of logarithms
Changing the base of a logarithm can be particularly useful if you need to evaluate a log problem without a calculator
- Choose the base such that you would know how to solve the problem from the equivalent exponent

This formula had more use when calculators were less advanced
- Some old calculators only had a button for logarithm of base 10
- To calculate $\log_{5} 7$ on these calculators you would have to enter
  - $\frac{\log_{10} 7}{\log_{10} 5}$

The formula can be useful when evaluating a logarithm where the two numbers are powers of a common number
- $\log_{4} 8 = \frac{\log_{2} 8}{\log_{2} 4} = \frac{3}{2}$
The formula can be useful when you are solving equations and two logarithms have different bases
- For example, if you have $\log_{3} k$ and $\log_{9} n$ within the same equation
  - You can rewrite $\log_{9} n$ as $\frac{\log_{3} n}{\log_{3} 9}$ which simplifies to $\frac{1}{2} \log_{3} n$
  - Or you can rewrite $\log_{3} k$ as $\frac{\log_{9} k}{\log_{9} 3}$ which simplifies to $2 \log_{9} k$
The formula also allows you to derive and use a formula for switching the numbers:

$\log_{a} x = \frac{1}{\log_{x} a}$

Using the fact that $\log_{x} x = 1$

Worked example

By choosing a suitable value for b, use the change of base law to find the value of $\log_{8} 32$ without using a calculator.

Note that 8 and 32 are both powers of 2, where 8 = 2³ and 32 = 2⁵.

Therefore we can choose b = 2 in the change of base formula.

$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$ :

$\log_{8} 32 = \frac{\log_{2} 32}{\log_{2} 8}$

If 2³ = 8 then log₂8 = 3 and if 2⁵ = 32 then log₂32 = 5.

$\frac{\log_{2} 32}{\log_{2} 8} = \frac{5}{3}$

$\log_{8} 32 = \frac{5}{3} = 1 \frac{2}{3}$