Properties of Logarithms | Edexcel IGCSE Further Maths Revision Notes 2019

Laws of Logarithms

What are the laws of logarithms?

Laws of logarithms allow you to simplify and manipulate expressions involving logarithms
- They can help with solving exponential and logarithmic equations
- The laws of logarithms are closely related to the laws of indices
You need to know the following laws, which are valid for $a, x, y > 0$ , $a \neq 1$ :
- $\log_{a} x y = \log_{a} x + \log_{a} y$
  - "log of a product is equal to the sum of the logs"
  - This relates to $a^{m} \times a^{n} = a^{m + n}$
- $\log_{a} \frac{x}{y} = \log_{a} x {- \log}_{a} y$
  - "log of a division is equal to the difference of the logs"
  - This relates to $a^{m} \div a^{n} = a^{m - n}$
- $\log_{a} x^{k} = k \log_{a} x$
  - "power in a log may be brought down as a multiplier in front of the log"
  - note that $\log_{a} x^{k} = \log_{a} (x^{k})$
  - This relates to $(a^{m})^{n} = a^{m n}$
- $\log_{a} a = 1$
  - This relates to $a^{1} = a$
- $\log_{a} 1 = 0$
  - This relates to $a^{0 = 1}$
Be careful
- With the first two laws the logs on the right-hand side must have the same base
  - Logs with different bases cannot be combined using those laws
- Also note the following (students often make these mistakes on the exam):
  - $\log_{a} (x + y)$ is not equal to $\log_{a} x + \log_{a} y$
  - $\log_{a} (x - y)$ is not equal to $\log_{a} x - \log_{a} y$

What are some other useful properties of logarithms?

You should also be familiar with the following properties of logarithms
- $\log_{a} a^{k} = k$
  - This can be derived from the third and fourth laws above
- $\log_{a} \frac{1}{x} = - \log_{a} x$
  - Because $\log_{a} \frac{1}{x} = \log_{a} (x^{- 1})$
  - Then use the third law above
- $\log_{a} (a^{x}) = a^{\log_{a} x} = x$
  - Because $\log_{a} x$ and $a^{x}$ are inverse functions
  - "log cancels exponential and exponential cancels log"
Also remember that $\ln x$ is another way of writing $\log_{e} x$
- All the laws and properties apply to $\ln x$ as well
- This includes $\ln (e^{x}) = e^{\ln x} = x$

Exam Tip

Make sure you know the laws and properties of logarithms
- They are not included on the exam formula sheet

Worked example

(a)

Write the expression

2 \log_{3} 4 - \log_{3} 2

in the form

\log_{3} p

, where

p

is an integer.

First use

\log_{a} x^{k} = k \log_{a} x

to rewrite

2 \log_{3} 4

\begin{array}{rcl} 2 \log_{3} 4 & = & \log_{3} 4^{2} \\ = & \log_{3} 16 \end{array}

Substitute that into the original expression

Then use

\log_{a} \frac{x}{y} = \log_{a} x {- \log}_{a} y

to simplify

\begin{array}{rcl} 2 \log_{3} 4 - \log_{3} 2 & = & \log_{3} 16 - \log_{3} 2 \\ = & \log_{3} \frac{16}{2} \\ = & \log_{3} 8 \end{array}

That's in the required form with

p = 8

\log_{3} 8

(b)

\begin{matrix}  \end{matrix}

Hence solve

2 \log_{3} 4 - \log_{3} 2 = - \log_{3} \frac{1}{x}

First use

\log_{a} x^{k} = k \log_{a} x

to rewrite

- \log_{3} \frac{1}{x}

Note that

{(\frac{1}{x})}^{- 1} = \frac{1}{(\frac{1}{x})} = x

\begin{array}{rcl} - \log_{3} \frac{1}{x} & = & \log_{3} {(\frac{1}{x})}^{- 1} \\ = & \log_{3} x \end{array}

Substitute that, and your answer from part (a), into the equation and solve for

x

\begin{array}{rcl} 2 \log_{3} 4 - \log_{3} 2 & = & - \log_{3} \frac{1}{x} \\ \log_{3} 8 & = & \log_{3} x \end{array}

x = 8

Change of Base

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(s) of the logarithm(s)
  - and then apply the logarithm laws
Changing the base can also allow a log problem to be solved without a calculator
- Choose a base that allows you to solve the problem using the equivalent exponent

How do I change the base of a logarithm?

The formula for changing the base of a logarithm is
- $\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$

This is on the formula sheet in the exam paper
- So you don't need to remember it
- But you do need to be able to use it
The change of base formula also leads to the following useful result:
- $\log_{a} b = \frac{1}{\log_{b} a}$
  - This is not on the formula sheet
  - But it can be derived from the change of base formula:

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

Exam Tip

Changing the base is a key skill
- Make sure you are confident with using it!
If you get stuck on a logarithm question, stop and think whether change of base would help
- And don't forget that the formula is on the exam formula sheet!

Worked example

Solve the equation

$\log_{4} x + \log_{32} x + \log_{2} x = \frac{34}{5}$

Show your working clearly.

$4$ and $32$ are both powers of 2

So use $\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$ with $b = 2$

This will make ALL the logarithms have a base of $2$

$\log_{4} x = \frac{\log_{2} x}{\log_{2} 4}$

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

Substitute into the equation

$\frac{\log_{2} x}{\log_{2} 4} + \frac{\log_{2} x}{\log_{2} 32} + \log_{2} x = \frac{34}{5}$

$\log_{2} 4 = 2$ because $2^{2} = 4$ (by the definition of a logarithm)

$\log_{2} 32 = 5$ because $2^{5} = 32$

Substitute into the equation and collect terms on the left-hand side

$\begin{array}{rcl} \frac{\log_{2} x}{2} + \frac{\log_{2} x}{5} + \log_{2} x & = & \frac{34}{5} \\ (\frac{1}{2} + \frac{1}{5} + 1) \log_{2} x & = & \frac{34}{5} \\ \frac{17}{10} \log_{2} x & = & \frac{34}{5} \end{array}$

Multiply both sides by $\frac{10}{17}$ to find the value of $\log_{2} x$

$\begin{array}{rcl} \log_{2} x & = & \frac{10}{17} \times \frac{34}{5} = 4 \end{array}$

By the definition of a logarithm, $\log_{2} x = 4$ is equivalent to $x = 2^{4}$

$x = 2^{4}$

$x = 16$

Properties of Logarithms (Edexcel IGCSE Further Maths)

Revision Note

Laws of Logarithms

What are the laws of logarithms?

What are some other useful properties of logarithms?

Exam Tip

Worked example

Change of Base

Why change the base of a logarithm?

How do I change the base of a logarithm?

Exam Tip

Worked example

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Author: Amber