A-Level Maths (Edexcel 9MA0) · Pure Year 1 · Chapter 14: Exponentials and logarithms · Lesson 3

Logarithms and the laws of logarithms

70 marks

Hints online, answers free with an account:
mathswariz.uk/worksheets/a-level-logarithms-and-the-laws-of-logarithms/

Date

Do Now

Quick questions to start.

A1
Needed today
Solve $3^{x} = \frac{1}{27}$ .
[2 marks]
Answer
A2
Last lesson
$V = 400 e^{- 0.2 t}$ . Find the rate of change of $V$ when $t = 3$ .
[2 marks]
Answer
A3
From chapter 1
Find the value of $1 6^{\frac{3}{4}}$ and of $1 6^{- \frac{1}{2}}$ .
[2 marks]
Answer
A4
GCSE Higher
Solve $x^{2} - 3 x - 10 = 0$ .
[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

(a) Write $5^{4} = 625$ as a logarithm. (b) Without a calculator, find $lo g_{4} \frac{1}{8}$ .

Hint

lo g_{a} n = x

means

a^{x} = n

. Write

4

and

\frac{1}{8}

as powers of

2

B1
(a) Write $lo g_{6} 216 = 3$ as a power. (b) Without a calculator, find $lo g_{8} \frac{1}{4}$ .
[3 marks]
Answer

Example 2

Show that $3 lo g_{2} 6 - lo g_{2} 27 = 3$ .

Hint Use the power law first, then the division law. Then ask:

2

to what power gives the number inside?

B2
Write $2 lo g_{5} 10 - lo g_{5} 4$ as a single logarithm, and hence find its value.
[3 marks]
Answer

Example 3

Write $lo g_{a} (\frac{x ^{3} y}{a ^{2}})$ in terms of $lo g_{a} x$ and $lo g_{a} y$ .

Hint Split it with the multiplication and division laws first.

y = y^{\frac{1}{2}}

, and

lo g_{a} a = 1

B3
Write $lo g_{a} (\frac{a x ^{2}}{y ^{4}})$ in terms of $lo g_{a} x$ and $lo g_{a} y$ .
[3 marks]
Answer

Example 4

Solve $lo g_{2} (x + 10) - lo g_{2} (x - 4) = 3$ .

Hint Use the division law to make one logarithm, then rewrite it as a power. Check the answer keeps

x - 4

positive.

B4
Solve $lo g_{3} (5 x + 8) - lo g_{3} (x - 4) = 2$ .
[3 marks]
Answer

Practice

A-Level

C1
Without a calculator, explain why $lo g_{3} 50$ lies between $3$ and $4$ . Then use your calculator to find $lo g_{3} 50$ to $3$ decimal places.
[3 marks]
Answer

C2
Solve $lo g_{x} (3 x + 10) = 2$ .
[3 marks]
Answer
C3
The positive numbers $a$ and $b$ satisfy $lo g_{2} a + lo g_{2} b = 5$ and $a - b = 4$ . Find $a$ and $b$ .
[5 marks]
Answer $a = \dots\dots$ $b = \dots\dots$
C4
By writing $lo g_{a} x = m$ , prove that $lo g_{a} (x^{k}) = k lo g_{a} x$ .
[3 marks]

Exam-style questions

Edexcel exam style

D1
An equation is $2 lo g_{4} (x - 2) = 1 + lo g_{4} x$ .
(a)
Show that the equation can be written as $x^{2} - 8 x + 4 = 0$ .
[4 marks]
(b)
Hence solve the equation, giving your answer in the form $a + b 3$ . Explain why the equation has only one solution.
[4 marks]
Total for D1: 8 marks

D2
Sam solves $lo g_{2} x + lo g_{2} (x + 2) = 3$ like this: " $lo g_{2} x + lo g_{2} (x + 2) = lo g_{2} (2 x + 2)$ , so $2 x + 2 = 8$ and $x = 3$ ."
Sam is wrong.
(a)
Show that $x = 3$ does not satisfy the equation.
[2 marks]
(b)
Explain Sam's mistake and solve the equation correctly.
[4 marks]
Answer
Total for D2: 6 marks
D3
The positive numbers $x$ and $y$ satisfy the simultaneous equations $2 lo g_{4} y - lo g_{4} x = 1 x + y = 15$
(a)
Show that $y^{2} = 4 x$ .
[3 marks]
(b)
Hence find the values of $x$ and $y$ .
[5 marks]
Answer $x = \dots\dots$ $y = \dots\dots$
Total for D3: 8 marks
D4
Given that $lo g_{a} 2 = p$ and $lo g_{a} 3 = q$ ,
(a)
write $lo g_{a} 72$ in terms of $p$ and $q$ .
[2 marks]
Answer
(b)
write $lo g_{a} (\frac{27}{4 a})$ in terms of $p$ and $q$ .
[3 marks]
Answer
(c)
Given also that $a = 6$ , show that $p + q = 1$ . Hence write $lo g_{6} 72$ in terms of $p$ only.
[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 the two values of $x$ for which $lo g_{2} x = lo g_{x} 16$ .
[6 marks]
Hint 1 · What to try
Call $lo g_{2} x = m$ , so $x = 2^{m}$ . Now write $lo g_{x} 16 = n$ as a power too.
Hint 2 · The first line
$x^{n} = 16$ gives $(2^{m})^{n} = 2^{4}$ , so $mn = 4$ . The equation says $m = n$ .
Hint 3 · The full method
$m^{2} = 4$ , so $m = 2$ or $m = - 2$ . Then $x = 2^{2} = 4$ or $x = 2^{- 2} = \frac{1}{4}$ . Both are positive and not $1$ , so both are allowed as a base.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can change between a logarithm and a power, and find a logarithm without a calculator.
I can use the three laws of logarithms, including the facts that a logarithm of its own base is 1 and a logarithm of 1 is 0.
I can solve equations with logarithms and reject a value that makes a logarithm undefined.
I can write logarithms in terms of given letters, such as p and q.

Answers and mark scheme

The worked answers and the full mark scheme are free with a Mathswariz account. Teachers and students can both sign up.

Get the answers