What is 6.6: Logarithmic graphs for modelling?

For y=ax^n, taking logarithms gives y= a+n x, so a plot of y against x has gradient n and intercept a.

How many marks are 6.6 questions worth?

6.6 practice drills on this page are worth 3–6 marks each; marks in a real exam vary by question.

Which past exam papers has 6.6 appeared in?

We have not yet indexed a verified past-paper appearance for 6.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.6

Use logarithmic graphs to estimate parameters in relationships of the form y = axⁿ and y = kbˣ, given data for x and y.

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Logarithmic graphs for modelling

Worked answers and methods for 6.6 on Edexcel A-level Maths 9MA0.

Explanation

For $y=ax^n$ , taking logarithms gives $\log y=\log a+n\log x$ , so a plot of $\log y$ against $\log x$ has gradient $n$ and intercept $\log a$ .
For $y=kb^x$ , $\log y=\log k+x\log b$ , so a plot of $\log y$ against $x$ has gradient $\log b$ and intercept $\log k$ .
Read two well-separated points from the fitted line, calculate its gradient and intercept, then undo the chosen logarithm base to recover $a$ , $k$ or $b$ .
The transformed coordinates and the logarithm base must be identified: confusing a log-log graph with a semi-log graph, or forgetting to exponentiate the intercept, gives incorrect parameters.

For

y=ax^n

, plotting

\log y

against

\log x

gives a straight line with gradient

n

and intercept

\log a

Worked example

For a model $y=ax^n$ , a fitted graph of $\log_{10}y$ against $\log_{10}x$ passes through $(0.2,0.65)$ and $(0.8,1.55)$ . Estimate $a$ and $n$ , then predict $y$ when $x=5$ .

1.The gradient is $n=(1.55-0.65)/(0.8-0.2)=1.5$ .
2.Using $(0.2,0.65)$ , the intercept is $\log_{10}a=0.65-1.5(0.2)=0.35$ , so $a=10^{0.35}=2.2387\ldots$ .
3.Thus at $x=5$ , $y=2.2387\ldots(5^{1.5})=25.030\ldots$ , giving about $25.0$ .

Answer: $n=1.5$ $a\approx2.24$ $y\approx25.0$

Common mistakes

Don't use the intercept itself as $a$ or $k$ instead of undoing the logarithm.
Don't plot $(\log x,\log y)$ for an exponential model $y=kb^x$ , which requires $(x,\log y)$ .

Exam tip

Name the transformed axes, then identify exactly which model parameter is the gradient and which is obtained from the intercept.

Worked practice

Tier 1 · Easy

A straight-line fit on base-

10

logarithmic axes has equation

\log_{10}y=0.4+1.7\log_{10}x

. State the corresponding model

y=ax^n

, giving

a

3

significant figures.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y=2.51x^{1.7}$	3
Notes
Comparing with $\log_{10}y=\log_{10}a+n\log_{10}x$ gives $n=1.7$ and $\log_{10}a=0.4$ . Hence $a=10^{0.4}=2.511\ldots$ , so $y=2.51x^{1.7}$ .

(3 marks)

Tier 2 · Standard

For a model

y=kb^x

, a fitted graph of

\ln y

against

x

passes through

(2,1.3)

and

(6,2.5)

. Estimate

k

and

b

3

significant figures, then predict

y

when

x=4

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$k\approx2.01$ $b\approx1.35$ $y\approx6.69$ when unrounded parameter values are carried; $y\approx6.68$ is also accepted if the stated $3$ s.f. values $k=2.01$ and $b=1.35$ are used	5
Notes
The straight-line form is $\ln y=\ln k+x\ln b$ . Its gradient is $(2.5-1.3)/(6-2)=0.3$ , so $b=e^{0.3}=1.3498\ldots$ . The intercept is $1.3-0.3(2)=0.7$ , so $k=e^{0.7}=2.0137\ldots$ . Prefer carrying these unrounded values: at $x=4$ , $\ln y=0.7+0.3(4)=1.9$ , so $y=e^{1.9}=6.6858\ldots\approx6.69$ . However, following the instruction to use the reported $3$ s.f. parameters gives $y=2.01(1.35)^4=6.676\ldots\approx6.68$ , which is also valid.

(5 marks)

Tier 3 · Hard

An exponential relationship

y=kb^x

is analysed by plotting

\ln y

against

x

. The fitted line goes through

(1,2.1)

and

(5,3.7)

. Find

k

and

b

3

significant figures, and estimate

y

x=3

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$k=5.47$ $b=1.49$ $y\approx18.2$	6
Notes
The straight-line form is $\ln y=\ln k+x\ln b$ . Its gradient is $(3.7-2.1)/(5-1)=0.4$ , so $\ln b=0.4$ and $b=e^{0.4}=1.4918\ldots$ . The intercept is $2.1-0.4(1)=1.7$ , so $k=e^{1.7}=5.4739\ldots$ . At $x=3$ , $\ln y=1.7+0.4(3)=2.9$ , giving $y=e^{2.9}=18.174\ldots$ .

(6 marks)

Tier 1 · Easy

A relationship is believed to have the form

y=kb^x

. State the quantities that should be plotted to obtain a straight line, and state what its gradient and vertical-axis intercept represent when natural logarithms are used.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Plot $\ln y$ against $x$ Gradient $=\ln b$ and vertical-axis intercept $=\ln k$	3
Notes
Taking logarithms gives $\ln y=\ln k+x\ln b$ . This is linear in $x$ , so plotting $\ln y$ vertically against $x$ horizontally gives gradient $\ln b$ and intercept $\ln k$ .

(3 marks)

Tier 2 · Standard

For a model

y=kb^x

, the fitted straight line has equation

\log_{10}y=0.52+0.28x

. Estimate

k

and

b

3

significant figures. Hence find the value of

x

for which

y=50

, giving

x

3

significant figures.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$k=3.31$ $b=1.91$ $x=4.21$	5
Notes
Comparing with $\log_{10}y=\log_{10}k+x\log_{10}b$ gives $k=10^{0.52}=3.31131\ldots$ and $b=10^{0.28}=1.90546\ldots$ . When $y=50$ , $\log_{10}50=0.52+0.28x$ , so $x=(\log_{10}50-0.52)/0.28=4.21061\ldots$ .

(5 marks)

Tier 3 · Hard

A relationship is modelled by

y=ax^n

. Because of a rescaled vertical axis, the fitted line is

\log_{10}(2y)=1.4-0.8\log_{10}x

. Find

a

and

n

, giving

a

3

significant figures. Hence estimate the value of

x

when

y=4

, giving

x

3

significant figures. Use unrounded values in your working.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$a=12.6$ $n=-0.8$ $x=4.18$	6
Notes
Since $\log_{10}(2y)=\log_{10}2+\log_{10}a+n\log_{10}x$ , the gradient gives $n=-0.8$ and the intercept gives $\log_{10}(2a)=1.4$ . Hence $a=10^{1.4}/2=12.5594\ldots$ . When $y=4$ , $\log_{10}8=1.4-0.8\log_{10}x$ , so $x=(10^{1.4}/8)^{1/0.8}=4.17963\ldots$ .

(6 marks)

Tier 2 · Standard

For a model

y=ax^n

, a fitted graph of

\log_{10}y

against

\log_{10}x

passes through

(-1,1.2)

and

(2,3.0)

. Find

n

and

a

, giving

a

3

significant figures. Find the value of

\log_{10}y

when

\log_{10}x=0.5

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$n=0.6$ $a=63.1$ $\log_{10}y=2.1$	5
Notes
The straight-line form is $\log_{10}y=\log_{10}a+n\log_{10}x$ . Its gradient is $n=(3.0-1.2)/(2-(-1))=0.6$ . Using $(-1,1.2)$ gives $\log_{10}a=1.2-0.6(-1)=1.8$ , so $a=10^{1.8}=63.0957\ldots=63.1$ to $3$ significant figures. When $\log_{10}x=0.5$ , $\log_{10}y=1.8+0.6(0.5)=2.1$ .

(5 marks)

Tier 3 · Hard

Three data pairs are

(x,y)=(4,16)

(9,54)

and

(16,128)

. Use logarithms to decide whether

y=ax^n

y=kb^x

fits all three pairs exactly. Determine the parameters of the chosen model and hence predict

y

when

x=25

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The power model fits $a=2$ and $n=\dfrac32$ $y=250$ when $x=25$	6
Notes
For a power model, the gradient between the first two log-log points is $\ln(54/16)/\ln(9/4)=\ln(27/8)/\ln(9/4)=3/2$ . Between the next two it is $\ln(128/54)/\ln(16/9)=\ln(64/27)/\ln(16/9)=3/2$ again, so the three log-log points are collinear. In contrast, the semi-log gradients $\ln(54/16)/(9-4)$ and $\ln(128/54)/(16-9)$ are unequal, so one exponential model does not fit all three pairs. Using $(4,16)$ in $y=ax^{3/2}$ gives $16=8a$ , so $a=2$ . Hence at $x=25$ , $y=2(25^{3/2})=2(125)=250$ .

(6 marks)

Tier 3 · Hard

A relationship has the form

y=ax^n

, where

x>0

. Doubling

x

multiplies

y

8

, and

y=20

when

x=2

. Determine

a

and

n

. Write the corresponding straight-line equation for a graph of

\log_{10}y

against

\log_{10}x

, and find

x

when

y=540

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=\dfrac52$ and $n=3$ $\log_{10}y=\log_{10}\!\left(\dfrac52\right)+3\log_{10}x$ $x=6$ when $y=540$	6
Notes
For a power model, $y(2x)/y(x)=2^n$ . The multiplier $8$ gives $2^n=8$ , so $n=3$ . Using $(x,y)=(2,20)$ gives $20=8a$ , hence $a=5/2$ . Taking base- $10$ logarithms gives the stated line. Finally, $540=(5/2)x^3$ , so $x^3=216$ and, since $x>0$ , $x=6$ .

(6 marks)

Q10

Tier 3 · Hard

10.

For positive

x

and

y

, a fitted graph of

\log_{10}x

against

\log_{10}y

has equation

\log_{10}x=0.4\log_{10}y-0.2

. Express the relationship in the form

y=ax^n

, giving

a

and

n

exactly. Hence predict the exact value of

y

when

x=40

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$y=\sqrt{10}\,x^{5/2}$ , so $a=\sqrt{10}$ and $n=\dfrac52$ $y=32000$ when $x=40$	5
Notes
Rearranging the fitted line gives $\log_{10}y=2.5\log_{10}x+0.5$ . Therefore $y=10^{0.5}x^{2.5}=\sqrt{10}\,x^{5/2}$ . At $x=40$ , $y=\sqrt{10}(40^2\sqrt{40})=\sqrt{10}(1600)(2\sqrt{10})=32000$ .

(5 marks)

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Other points in 6 Exponentials and logarithms

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