A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 14: Exponentials and logarithms · Lesson 4

Solving equations with logarithms, and natural logarithms

70 marks

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Quick questions to start.

A1
Last lesson
Solve $lo g_{3} (2 x - 1) = 2$ .
[2 marks]
Answer
A2
Needed today
Write $2^{6} = 64$ as a logarithm. Find $lo g_{3} 243$ .
[2 marks]
Answer
A3
GCSE Higher
Solve $u^{2} - 5 u - 14 = 0$ .
[2 marks]
Answer
A4
This chapter
$P = 250 e^{0.04 t}$ . Find $P$ when $t = 10$ , to the nearest whole number.
[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

Solve, giving each answer to $3$ significant figures. (a) $4^{x} = 30$ (b) $3^{2 x + 1} = 50$

Hint

a^{x} = b

means

x = lo g_{a} b

. In (b), first find what the power

2 x + 1

must be.

B1
Solve, giving each answer to $3$ significant figures. (a) $6^{x} = 45$ (b) $5^{3 x - 2} = 80$
[3 marks]
Answer

Example 2

Solve $2^{2 x} - 9 (2^{x}) + 14 = 0$ , giving answers to $3$ significant figures where needed.

Hint

2^{2 x} = (2^{x})^{2}

, so this is a quadratic in

2^{x}

. Let

y = 2^{x}

B2
Solve $3^{2 x} - 13 (3^{x}) + 30 = 0$ , giving answers to $3$ significant figures where needed.
[3 marks]
Answer

Example 3

Solve $5^{x} = 3^{x + 2}$ , giving your answer to $3$ significant figures.

Hint Take

ln

of both sides, then use the power law. Collect the terms in

x

on one side and factorise.

B3
Solve $7^{x} = 2^{x + 3}$ , giving your answer to $3$ significant figures.
[3 marks]
Answer

Example 4

Solve, giving exact answers. (a) $e^{3 x - 1} = 8$ (b) $3 ln (2 x) + 1 = 7$

Hint

ln

undoes

e

, and

e

undoes

ln

. Get the

e

or the

ln

on its own first.

B4
Solve, giving exact answers. (a) $e^{2 x + 5} = 9$ (b) $ln (5 x - 2) = 3$
[3 marks]
Answer

Practice

A-Level

C1
Solve $e^{2 x} - 3 e^{x} - 10 = 0$ , giving your answer in exact form.
[3 marks]
Answer

C2
Solve $2 e^{x} - 5 = 3 e^{- x}$ , giving your answer in exact form.
[4 marks]
Answer
C3
Solve $(ln x)^{2} + ln x - 6 = 0$ , giving your answers in exact form.
[3 marks]
Answer
C4
On one set of axes, sketch $y = e^{x}$ and $y = ln x$ . Show the line each graph is a reflection of the other in, the point where each graph crosses an axis, and the asymptote of each graph.
[4 marks]

Exam-style questions

Edexcel exam style

D1
The mass, $m$ grams, of a radioactive substance $t$ days after it is made is modelled by $m = 80 e^{- 0.03 t}$ . A second substance has mass $m = 50 e^{- 0.01 t}$ grams $t$ days after it is made. Both are made at the same time.
(a)
Find, to the nearest day, the time it takes the first substance to fall to $20$ grams.
[3 marks]
Answer
(b)
Show that the two substances have the same mass when $t = 50 ln 1.6$ , and find this time to $1$ decimal place.
[4 marks]
(c)
Hence state the times for which the first substance has the greater mass.
[1 mark]
Answer
Total for D1: 8 marks

D2
Leo says: " $ln (x^{2}) = 2 ln x$ , so $ln (x^{2}) = 4$ means $ln x = 2$ , and the only solution is $x = e^{2}$ ."
Leo is wrong.
(a)
Show that $x = - e^{2}$ also satisfies $ln (x^{2}) = 4$ .
[2 marks]
(b)
Explain why Leo's step loses a solution. Then solve $ln (x^{2}) = 2 ln (x + 3)$ .
[4 marks]
Answer
Total for D2: 6 marks
D3
An equation is $4^{x + 1} - 17 (2^{x}) + 4 = 0$ .
(a)
Show that, with $y = 2^{x}$ , the equation can be written as $4 y^{2} - 17 y + 4 = 0$ .
[3 marks]
(b)
Hence solve the equation, without a calculator.
[3 marks]
Answer
(c)
Hence solve $4^{x + 1} - 17 (2^{x}) + 4 < 0$ .
[2 marks]
Answer
Total for D3: 8 marks
D4
This question is about the equation $ln (2 x + 3) + ln (x - 1) = 2 ln (x + 1)$ .
(a)
Solve the equation, giving your answer in the form $\frac{a + b}{2}$ . Explain why it has only one solution.
[5 marks]
Answer
(b)
Hence solve $ln (2 e^{y} + 3) + ln (e^{y} - 1) = 2 ln (e^{y} + 1)$ . Give your answer exactly, and to $3$ significant figures.
[3 marks]
Answer
Total for D4: 8 marks

Extension

Stretch

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

E1
Find, in exact form, the two values of $x$ for which $x^{l n x} = e^{4} x^{3}$ .
[6 marks]
Hint 1 · What to try
Take $ln$ of both sides. The power law turns $ln (x^{l n x})$ into $ln x \times ln x$ .
Hint 2 · The first line
The right side becomes $4 + 3 ln x$ , so $(ln x)^{2} = 4 + 3 ln x$ . Let $u = ln x$ .
Hint 3 · The full method
$u^{2} - 3 u - 4 = 0$ , so $(u - 4) (u + 1) = 0$ . Then $x = e^{4}$ or $x = e^{- 1}$ . Both are positive, so both work.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can solve an equation with the unknown in a power by taking logarithms.
I can solve a quadratic in a power such as 2 to the x or e to the x, and reject a power that would be negative.
I can solve equations with e and ln and give exact answers.
I can use an exponential model to find when a quantity reaches a value.

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