Properties of Logarithms (Edexcel IGCSE Further Pure Maths): Revision Note

Exam code: 4PM1

Amber

Written by: Amber

Reviewed by: Dan Finlay

Updated on

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>0a1:

    • logaxy= logax+ logay

      • "log of a product is equal to the sum of the logs"

      • This relates to am× an=am+n

    • logaxy= logax  logay 

      • "log of a division is equal to the difference of the logs"

      • This relates to am÷ an=amn

    • logaxk= klogax 

      • "power in a log may be brought down as a multiplier in front of the log"

      • note that  logaxk=loga(xk)

      • This relates to (am)n=amn

    • logaa=1

      • This relates to  a1=a

    • loga1=0

      • This relates to  a0=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):

      • loga(x+y) is not equal to logax+logay

      • loga(xy) is not equal to logaxlogay

What are some other useful properties of logarithms?

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

    • logaak= k

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

    • loga1x=logax

      • Because loga1x=loga(x1)

      • Then use the third law above

    • loga(ax)=alogax= x

      • Because logax and ax are inverse functions

      • "log cancels exponential and exponential cancels log"

  • Also remember that lnx is another way of writing logex

    • All the laws and properties apply to lnx as well

    • This includes  ln(ex)=elnx=x

Examiner Tips and Tricks

  • 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  2log34log32  in the form  log3p,  where p is an integer.

First use  logaxk= klogax  to rewrite 2log34

 2log34=log342=log316

Substitute that into the original expression
Then use  logaxy= logax  logay  to simplify

2log34log32=log316log32=log3162=log38
That's in the required form with p=8

log38
 

(b) Hence solve   2log3 4log32=log31x.

First use  logaxk= klogax  to rewrite  log31x

Note that  (1x)1=1(1x)=x

log31x=log3(1x)1=log3x

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

 2log3 4log32=log31xlog38=log3x

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

    • logax= logbxlogba

  • 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:

    • logab=1logba

      • This is not on the formula sheet

      • But it can be derived from the change of base formula:

logab=logbblogba=1logba

Examiner Tips and Tricks

  • 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

log4x+log32x+log2x=345

Show your working clearly.

4 and 32 are both powers of 2

So use  logax= logbxlogba  with  b=2

This will make ALL the logarithms have a base of 2

log4x= log2xlog24

log32x= log2xlog232


Substitute into the equation

log2xlog24+log2xlog232+log2x=345

log24=2  because  22=4 (by the definition of a logarithm)

log232=5  because  25=32

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


log2x2+log2x5+log2x=345(12+15+1)log2x=3451710log2x=345

Multiply both sides by 1017 to find the value of  log2x


log2x=1017×345=4


By the definition of a logarithm,  log2x=4  is equivalent to  x=24   

x=24

x=16

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

Amber

Author: Amber

Expertise: Maths Content Creator

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.