Properties of Logarithms (Edexcel IGCSE Further Maths)

Laws of Logarithms

Spec point — 1B

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 $a,x,y>0$, $a\neq1$:

  • $\mathrm{log}_{a}xy=\mathrm{log}_{a}x+\mathrm{log}_{a}y$

    • "log of a product is equal to the sum of the logs"
    • This relates to $a^{m}\times a^{n}=a^{m+n}$

  • $\mathrm{log}_{a}\frac{x}{y}=\mathrm{log}_{a}x-\mathrm{log}_{a}y$

    • "log of a division is equal to the difference of the logs"
    • This relates to $a^{m}\div a^{n}=a^{m-n}$

  • $\mathrm{log}_{a}x^{k}=k\mathrm{log}_{a}x$

    • "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 $(a^{m})^{n}=a^{mn}$

  • $\mathrm{log}_{a}a=1$

    • This relates to $a^{1}=a$

  • $\mathrm{log}_{a}1=0$

    • 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 $\mathrm{log}_{a}x+\mathrm{log}_{a}y$
    • $log_{a} ( x - y )$ is not equal to $\mathrm{log}_{a}x-\mathrm{log}_{a}y$

What are some other useful properties of logarithms?

• You should also be familiar with the following properties of logarithms

  • $\mathrm{log}_{a}a^{k}=k$

    • This can be derived from the third and fourth laws above

  • $\mathrm{log}_{a}\frac{1}{x}=-\mathrm{log}_{a}x$

    • 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 $\mathrm{log}_{a}x$ and $a^{x}$ are inverse functions
    • "log cancels exponential and exponential cancels log"

• Also remember that $\mathrm{ln}x$ is another way of writing $\mathrm{log}_{e}x$

  • All the laws and properties apply to $\mathrm{ln}x$ as well
  • This includes $\ln ( \mathrm{e}^{x ) = \mathrm{e}^{\ln } x} = x$

Exam Hint

• 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 $2\mathrm{log}_{3}4-\mathrm{log}_{3}2$ in the form $\mathrm{log}_{3}p$, where $p$ is an integer.

First use $\mathrm{log}_{a}x^{k}=k\mathrm{log}_{a}x$ to rewrite $2\mathrm{log}_{3}4$

$\begin{gathered} 2 log_{3} 4 & = log_{3} 4^{2} \\ = log_{3} 16 \end{gathered}$

Substitute that into the original expression Then use $\mathrm{log}_{a}\frac{x}{y}=\mathrm{log}_{a}x-\mathrm{log}_{a}y$ to simplify

$\begin{gathered} 2 log_{3} 4 - log_{3} 2 & = log_{3} 16 - log_{3} 2 \\ = log_{3} \frac{16}{2} \\ = log_{3} 8 \end{gathered}$ That's in the required form with $p=8$

$\mathrm{log}_{3}8$

(b) Hence solve $2\mathrm{log}_{3}4-\mathrm{log}_{3}2=-\mathrm{log}_{3}\frac{1}{x}$.

First use $\mathrm{log}_{a}x^{k}=k\mathrm{log}_{a}x$ to rewrite $-\mathrm{log}_{3}\frac{1}{x}$

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 $x$

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

$x=8$

Change of Base

Spec point — 1B

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

  • $\mathrm{log}_{a}x=\frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$

• 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:

  • $\mathrm{log}_{a}b=\frac{1}{\mathrm{log}_{b}a}$

    • This is not on the formula sheet
    • But it can be derived from the change of base formula:

$\mathrm{log}_{a}b=\frac{\mathrm{log}_{b}b}{\mathrm{log}_{b}a}=\frac{1}{\mathrm{log}_{b}a}$

Exam Hint

• 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

$\mathrm{log}_{4}x+\mathrm{log}_{32}x+\mathrm{log}_{2}x=\frac{34}{5}$

Show your working clearly.

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

So use $\mathrm{log}_{a}x=\frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$ with $b=2$

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

$\mathrm{log}_{4}x=\frac{\mathrm{log}_{2}x}{\mathrm{log}_{2}4}$

$\mathrm{log}_{32}x=\frac{\mathrm{log}_{2}x}{\mathrm{log}_{2}32}$

Substitute into the equation

$\frac{\mathrm{log}_{2}x}{\mathrm{log}_{2}4}+\frac{\mathrm{log}_{2}x}{\mathrm{log}_{2}32}+\mathrm{log}_{2}x=\frac{34}{5}$

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

$\mathrm{log}_{2}32=5$ because $2^{5}=32$

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

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

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

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

By the definition of a logarithm, $\mathrm{log}_{2}x=4$ is equivalent to $x=2^{4}$* *

$x=2^{4}$

$x=16$