A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 14: Exponentials and logarithms · Lesson 5

Logarithms and non-linear data

70 marks

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Date

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Quick questions to start.

A1
Last lesson
Solve $1 0^{x} = 250$ , giving your answer to $3$ significant figures.
[2 marks]
Answer
A2
This chapter
Write $lo g 5 + 3 lo g x$ as a single logarithm.
[2 marks]
Answer
A3
GCSE Higher
A straight line passes through $(0, 1.4)$ and $(5, 2.9)$ . Find its equation.
[2 marks]
Answer
A4
Needed today
Write $ln (6 e^{2 x})$ in the form $p + q x$ , where $p$ and $q$ are exact.
[2 marks]
Answer

Example, then your turn

A-Level

Your teacher works through each example with you. Then try the one beside it.

Example 1

Which graph of logs is a straight line? Give its gradient and vertical intercept to $3$ s.f. (a) $y = 5 x^{3}$ (b) $N = 40 \times 1. 3^{t}$

Hint Here

lo g

means

lo g_{10}

. Take

lo g

of both sides and use the laws. In (b) the power law brings down

t

, not

lo g t

B1
Which graph of logs is a straight line? Give its gradient and vertical intercept to $3$ s.f. (a) $T = 12 x^{2.5}$ (b) $P = 320 \times 0.8 5^{t}$
[3 marks]
Answer

Example 2

$y = a b^{x}$ . The graph of $lo g y$ against $x$ is a straight line through $(0, 0.8)$ and $(4, 2.0)$ . Find $a$ and $b$ to $3$ s.f. What does $a$ mean?

Not drawn accurately

Hint

lo g y = lo g a + x lo g b

. So the gradient is

lo g b

and the intercept is

lo g a

B2
A population is $P = a b^{t}$ after $t$ years. The graph of $lo g P$ against $t$ is a straight line through $(0, 1.2)$ and $(5, 0.2)$ . Find $a$ and $b$ to $3$ s.f. What does $a$ mean?
[3 marks]
Not drawn accurately
Answer

Example 3

$y = a x^{n}$ . The graph of $lo g y$ against $lo g x$ is a straight line through $(0, 0.5)$ and $(2, 3.5)$ . Find $n$ , and $a$ to $3$ s.f.

Not drawn accurately

Hint

lo g y = lo g a + n lo g x

. So the gradient is

n

and the intercept is

lo g a

B3
$y = a x^{n}$ . The graph of $lo g y$ against $lo g x$ is a straight line through $(1, 2.3)$ and $(3, 1.3)$ . Find $n$ , and $a$ to $3$ s.f.
[3 marks]
Not drawn accurately
Answer

Practice

A-Level

C1
The data in the table follow a model of the form $y = a b^{x}$ . By finding $lo g y$ for the first and last points, find $a$ and $b$ to $2$ significant figures.
[4 marks]
x 1 3 5
y 6.0 13.5 30.4
Answer

x	1	3	5
y	6.0	13.5	30.4

C2
Sam has data that follow $y = 5 \times 3^{x}$ . He says: "If I plot $lo g y$ against $lo g x$ , the points will lie on a straight line."
Explain why Sam is wrong. State which graph is a straight line, and give its gradient and intercept to $3$ significant figures.
[4 marks]
Answer
C3
A quantity is modelled by $y = A e^{k t}$ . The graph of $ln y$ against $t$ is a straight line with gradient $0.4$ that meets the vertical axis at $1.5$ . Find $k$ , and find $A$ to $3$ significant figures. Hence find $y$ when $t = 2$ .
[4 marks]
Not drawn accurately
Answer

Exam-style questions

Edexcel exam style

D1
The value, $V$ pounds, of a painting $t$ years after it is bought is shown in the table. The value is modelled by $V = a b^{t}$ , where $a$ and $b$ are constants.
t 0 10 20 30
V 1200 2150 3850 6890
(a)
Show that $lo g V = lo g a + t lo g b$ .
[2 marks]
(b)
Find $lo g V$ , to $2$ decimal places, for each value of $t$ . Using the values of $lo g V$ when $t = 0$ and $t = 30$ , find $a$ and $b$ to $3$ significant figures. Hence say what the value of $b$ tells you about the value of the painting.
[6 marks]
Answer
Total for D1: 8 marks

t	0	10	20	30
V	1200	2150	3850	6890

D2
A pendulum of length $L$ m has period $T$ s, where $T = a L^{n}$ . The graph of $lo g T$ against $lo g L$ is a straight line through $(- 1, - 0.2)$ and $(0.4, 0.5)$ . Kai says: "So $lo g a = - 0.2$ and $a = 0.631$ ."
Kai is wrong.
Not drawn accurately
(a)
Explain Kai's mistake. Find the correct values of $n$ and $a$ , giving $a$ to $3$ significant figures.
[4 marks]
Answer
(b)
Hence find, to $3$ significant figures, the length of a pendulum with a period of $3$ seconds, and the period of a pendulum $0.5$ m long.
[4 marks]
Answer
Total for D2: 8 marks
D3
The mass, $m$ mg, of a drug in a patient's blood $t$ hours after it is given is modelled by $m = A e^{- k t}$ , where $A$ and $k$ are positive constants. The graph of $ln m$ against $t$ is a straight line through $(2, 3.1)$ and $(8, 1.9)$ .
Not drawn accurately
(a)
Show that $ln m = ln A - k t$ . Find $k$ , and find $A$ to $3$ significant figures.
[6 marks]
Answer
(b)
Hence find the time when the mass first falls below $5$ mg, to $3$ significant figures.
[3 marks]
Answer
Total for D3: 9 marks
D4
An experiment gives the results in the table. The results are thought to follow either $y = a x^{n}$ or $y = a b^{x}$ .
x 1 2 4 8
y 2.5 4.35 7.58 13.2
(a)
Find $lo g x$ and $lo g y$ for each pair, to $2$ decimal places. Use them to decide which model fits, giving a reason.
[4 marks]
Answer
(b)
Find $n$ and $a$ to $2$ significant figures.
[3 marks]
Answer
(c)
Hence estimate $y$ when $x = 6$ and when $x = 50$ , to $3$ significant figures. Say which estimate is more reliable, and why.
[2 marks]
Answer
Total for D4: 9 marks

x	1	2	4	8
y	2.5	4.35	7.58	13.2

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
$y = a x^{n}$ , and the graph of $lo g y$ against $lo g x$ passes through $(lo g 4, lo g 24)$ and $(lo g 9, lo g 81)$ . Without a calculator, find $a$ and $n$ exactly, and find $x$ when $y = 375$ .
[7 marks]
Hint 1 · What to try
Both points are on the curve: $24 = a \times 4^{n}$ and $81 = a \times 9^{n}$ . Divide to remove $a$ .
Hint 2 · The first line
$\frac{81}{24} = (\frac{9}{4})^{n}$ , which is $(\frac{3}{2})^{3} = (\frac{3}{2})^{2 n}$ .
Hint 3 · The full method
So $n = \frac{3}{2}$ and $24 = 8 a$ . Then solve $3 x^{3/2} = 375$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can make y = ax^n a straight line: log y against log x.
I can make y = ab^x a straight line: log y against x.
I can find a model's constants from a graph of logs.
I can use a model and judge its predictions.

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