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Transforming Relationships to Linear Form (CIE IGCSE Additional Maths)

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Transforming Relationships in the Form y=ax^n

How do I use logarithms to linearise a graph in the form y = axn?

  • Logarithms can be used to linearise graphs of power functions 
  • Suppose y equals a x to the power of n
    • You can take logarithms of both sides
      • ln y equals ln left parenthesis a x to the power of n right parenthesis
    • You can split the right hand side into the sum of two logarithms
      • ln y equals ln a plus ln left parenthesis x to the power of n right parenthesis
    • You can bring down the power in the final term
      • ln y equals ln a plus n ln x
  • ln y equals ln a plus n ln x is in linear form Y equals m X plus c
    • Y equals ln y
    • X equals ln x
    • m equals n
    • c equals ln a

How can I use linearised form to find the unknown constants?

  • After linearising the function it will be in the form ln y equals ln a plus n ln x
    • n is the gradient of the straight line graph
    • ln space a is the y-intercept of the straight line graph
  • Once you know the value of the gradient of the straight line graph this is the value of n
  • You will need to find the value of a by solving the equation ln a space equals space c

Examiner Tip

  • You may need to leave your answer in exact form, especially in the non-calculator paper
    • e.g. to solve ln a = 2 the answer will be a = e2

Worked example

The heights, h metres, and the amount of time spent sleeping, t hours, of a group of young giraffes can be modelled using t space equals space a h to the power of b, where a and b are constants. 

The graph of ln space h against ln space t is a straight line passing through the points (1, -0.9) and (4, -4.5).

Find the values of a and b, giving your answers in exact form. 

Find the gradient of the straight line between the two coordinates.

m space equals space fraction numerator negative 4.5 space minus space minus 0.9 over denominator 4 minus 1 end fraction space equals space minus fraction numerator 3.6 over denominator 3 end fraction equals negative 1.2

Substitute m space equals space minus 1.2, x space equals space ln space h and y space equals space ln space t into the equation of a straight line (y space equals space m x space plus space c).

ln space t space equals space minus 1.2 ln space h space plus space c 

Substitute either coordinate in and rearrange to find c.

table row cell negative 0.9 space end cell equals cell space minus 1.2 open parentheses 1 close parentheses space plus space c end cell row cell c space end cell equals cell space minus 0.9 space plus space 1.2 end cell row cell c space end cell equals cell space 0.3 end cell end table

Compare this to the linearised form of t space equals space a h to the power of b.
Take logarithms of both sides first and rearrange if you can't remember the correct form. 

ln space t space equals space ln space open parentheses a h to the power of b close parentheses ln space t space equals space ln a space plus space b ln space h 

So b is the value in front of ln space h and ln space a space equals space 0.3.
Solve ln space a space equals space 0.3 by taking straight e of both sides.

table row cell ln space a space end cell equals cell space 0.3 end cell row cell straight e to the power of ln space a space end exponent end cell equals cell space straight e to the power of 0.3 end exponent end cell row cell a space end cell equals cell space straight e to the power of 0.3 end exponent end cell end table

bold italic a bold space bold equals bold space bold e to the power of bold 0 bold. bold 3 end exponent bold space bold comma bold space bold italic b bold space bold equals bold space bold minus bold 1 bold. bold 2

Transforming Relationships in the Form y=Ab^x

How do I use logarithms to linearise a graph in the form y = A(bx)?

  • Logarithms can be used to linearise graphs of exponential functions 
  • Suppose y equals A b to the power of x
    • You can take logarithms of both sides
      • log space y equals log space left parenthesis A b to the power of x right parenthesis
    • You can split the right hand side into the sum of two logarithms
      • log space y equals log space A plus log space left parenthesis b to the power of x right parenthesis
    • You can bring down the power in the final term
      • log space y space equals space log space A plus space x log space b
  • log space y space equals space log space A plus space x log space b is in linear form Y equals m X plus c
    • Y space equals space log space y
    • X space equals space x
    • m equals log space b
    • c space equals space log space A

How can I use linearised form to find the unknown constants?

  • After linearising the function it will be in the form log space y space equals space log space A plus space x log space b
    • log space b is the gradient of the straight line graph open parentheses m close parentheses
    • log space A is the y-intercept of the straight line graph open parentheses c close parentheses
  • You will need to find the value of A by solving the equation log space A space equals space c
    • The value of c will either be given or will need to be found
  • You will need to find the value of b by solving the equation log space b space equals space m
    • The value of m will either be given or will need to be found

Examiner Tip

  • Unless the question specifies, you can choose whether to use ln, lg or log
  • Remember you will need to solve the equation at the end
    • If using lg, solve by taking 10 to the power of each side
    • If using ln, solve by taking e to the power of each side

Worked example

Variables x and y are such that when lg space y is plotted against x, a straight line passing through the points left parenthesis 2 comma space 5 right parenthesis and left parenthesis 5 comma space 8 right parenthesis is obtained.

Show that y space equals space A space cross times space b to the power of x where A and bare constants to be found. 

Find the gradient of the straight line between the two coordinates.

m space equals space fraction numerator 8 space minus space 5 over denominator 5 space minus space 2 end fraction space equals space 3 over 3 space equals space 1

Substitute m space equals space 1, X space equals space x and Y space equals space lg space y into the equation of a straight line (Y space equals space m X space plus space c).

lg space y space equals space x space plus space c 

Substitute either coordinate in and rearrange to find c.

table row cell 5 space end cell equals cell space 2 space plus space c end cell row cell c space end cell equals cell space 3 end cell end table

lg space y space equals space x space plus space 3

Linearise y space equals space A cross times b to the power of x.
Take logarithms of both sides first and rearrange if you can't remember the correct form. 
Take logarithms of both sides. 

lg space y equals space lg space open parentheses A b to the power of x close parentheses 

Split the right-hand side into the sum of two logarithms.

lg space y space equals space lg space A space plus space lg space b to the power of x

Bring down the power in the final term. 

lg space y space equals space lg space A space plus space x lg space b

Compare this to the equation of the line.

lg space b space equals space 1 and lg space A space equals space 3 

Solve lg space b space equals space 1 by raising 10 to the power of both sides.

table row cell lg space b space end cell equals cell space 1 end cell row cell 10 to the power of lg space b end exponent space end cell equals cell space 10 to the power of 1 end cell row cell b space end cell equals cell space 10 end cell end table

Solve lg space A space equals space 3 by raising 10 to the power of both sides.

table row cell lg space A space end cell equals cell space 3 end cell row cell 10 to the power of lg space A end exponent space end cell equals cell space 10 cubed end cell row cell A space end cell equals cell space 1000 end cell end table

bold italic y bold space bold equals bold space bold 1000 bold space bold cross times bold space bold 10 to the power of bold italic x

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    Amber

    Author: Amber

    Expertise: Maths

    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.