What is 6.4: Laws of logarithms?

The laws convert between one logarithm and several: a product becomes a sum, a quotient a difference, and a power a coefficient, so ^3√(y)/z^2=3 x+1/2 y-2 z.

How many marks are 6.4 questions worth?

6.4 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 6.4 appeared in?

We have not yet indexed a verified past-paper appearance for 6.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

6.4

Understand and use the laws of logarithms: logₐx + logₐy = logₐ(xy); logₐx − logₐy = logₐ(x/y); k logₐx = logₐxᵏ (including, for example, k = −1 and k = −½).

Draft — not yet indexed

Laws of logarithms

Worked answers and methods for 6.4 on Edexcel A-level Maths 9MA0.

Definition

Specification 6.4: understand and use the laws of logarithms: $\log_a x+\log_a y=\log_a(xy)$ ; $\log_a x-\log_a y=\log_a\left(\dfrac xy\right)$ ; $k\log_a x=\log_a x^k$ , including for example $k=-1$ and $k=-\frac12$ .

Explanation

The laws convert between one logarithm and several: a product becomes a sum, a quotient a difference, and a power a coefficient, so $\ln\dfrac{x^3\sqrt y}{z^2}=3\ln x+\frac12\ln y-2\ln z$ .
Read backwards they collect an equation into a single logarithm, $\log_2(x+6)+\log_2(x-2)=3$ becoming $\log_2\big((x+6)(x-2)\big)=3$ , which is then removed by writing the argument equal to $2^3$ ; “taking logs” of $y=ab^x$ gives $\log y=\log a+x\log b$ , a straight line in $x$ whose intercept and gradient give $a$ and $b$ .
What the laws do not do: $\log(x+y)$ is not $\log x+\log y$ , and $\log x\cdot\log y$ is not $\log(xy)$ .
Because a logarithm needs a positive argument, every solution is checked in the original equation and any that makes an argument zero or negative is rejected with that reason. $\log_a a=1$ and $\log_a1=0$ .

Worked example

Solve

\log_2(x+6)+\log_2(x-2)=3

(5)

(Total for Question 1 is 5 marks)

Working

1.Combine using the addition law: $\log_2\big((x+6)(x-2)\big)=3$ .
2.Remove the logarithm: $(x+6)(x-2)=2^3=8$ , so $x^2+4x-12=8$ and $x^2+4x-20=0$ .
3.Solve: $x=\dfrac{-4\pm\sqrt{16+80}}{2}=-2\pm2\sqrt6$ .
4.The argument $x-2$ must be positive, so reject $x=-2-2\sqrt6$ ; the solution is $x=-2+2\sqrt6$ .

Answer: $x=-2+2\sqrt6$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Uses the addition law to obtain $\log_2\big((x+6)(x-2)\big)=3$	M1
	Uses $\log_2A=3\Rightarrow A=2^3$ to remove the logarithm	M1
	$x^2+4x-20=0$	A1
	Solves their three-term quadratic by a valid method	M1
	$x=-2+2\sqrt6$ only, with $-2-2\sqrt6$ rejected	A1
Notes
M1: Evidence of a correct logarithm law
A1: oe three-term quadratic
A1: cso

(5 marks)

Common mistakes

Don't split $\log(x+y)$ into $\log x+\log y$ , although there is no logarithm law for a sum.
Don't combine logarithms without retaining the condition that every original argument is positive.

Mark-scheme wording to use

“M1 for evidence of a correct logarithm law. M1 for combining the logarithms correctly into one logarithmic equation or equal-log form. B1 for using log_2(A) = 3 to obtain A = 2^3 = 8. A1* for completing the proof to x^3 + 4x^2 - 4x + 48 = 0 with no errors” — Pearson 9MA0, Specimen Paper 1, Q5(b)
“M1: Correct strategy to get a numerical expression or value for a or b e.g. a = 10^0.81 or b = 10^0.0054 … A1: Correct value for a or b. Allow 3 sf for this mark so allow a = awrt 6.46 or b = awrt 1.01. … M1: Correct strategy to get a numerical expression or value for a and b e.g. a = 10^0.81 and b = 10^0.0054 … A1: Correct values. This requires a = awrt 6.457 and b = awrt 1.013 for this mark.” — Pearson 9MA0, June 2024 Paper 2, Q13(a)
“M1: Take logarithms of both sides and show use of the addition law. A1: Use the power law to obtain log_10(N)=b log_10(T)+log_10(a) and state m=b and c=log_10(a)” — Pearson 9MA0, Sample Paper 1, Q12(a)

What the examiner wants

The laws applied as written, and a check that the arguments stay positive. The June 2024 Paper 1 report says attempts at using logarithms were mostly unsuccessful because the laws were not adhered to, and that candidates whose model gave the logarithm of a negative value should have taken it as a sign of an earlier error rather than forcing a positive answer.

Exam tip

For “single logarithm”, apply powers first, then combine products and quotients while keeping domain restrictions.

Worked practice

Tier 1 · Easy

Write

\ln12+\ln3-\ln2

as a single logarithm.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\ln18$	2
Notes
Use product and quotient laws: $\ln12+\ln3-\ln2=\ln(12\times3/2)=\ln18$ .

(2 marks)

Tier 2 · Standard

Solve

2\ln(x+1)-\ln x=\ln8

, giving all solutions allowed by the original equation.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$x=3-2\sqrt2$ or $x=3+2\sqrt2$	5
Notes
The logarithms require $x>0$ . Combine them to get $\ln\!\left((x+1)^2/x\right)=\ln8$ , so $(x+1)^2=8x$ . Hence $x^2-6x+1=0$ , giving $x=3\pm2\sqrt2$ . Both values are positive, so both satisfy the domain and are valid.

(5 marks)

Tier 3 · Hard

Solve

\log_3(x-1)+\log_3(x+3)=2

, checking the domain of the original equation.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x=-1+\sqrt{13}$	5
Notes
The original arguments require $x>1$ . Combine the logarithms: $\log_3((x-1)(x+3))=2$ , so $(x-1)(x+3)=9$ . Hence $x^2+2x-12=0$ , giving $x=-1\pm\sqrt{13}$ . Only $-1+\sqrt{13}>1$ , so the other algebraic root is rejected.

(5 marks)

Tier 1 · Easy

Given that

p>0

and

q>0

, express

3\log_a p-\dfrac12\log_a q

as a single logarithm.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\log_a\!\left(\dfrac{p^3}{\sqrt q}\right)$	2
Notes
Use the power law first: $3\log_a p=\log_a(p^3)$ and $\tfrac12\log_a q=\log_a(\sqrt q)$ . The subtraction law then gives $\log_a(p^3/\sqrt q)$ .

(2 marks)

Tier 2 · Standard

Given that

\log_a2=p

and

\log_a3=q

, express

\log_{\sqrt a}72

in terms of

p

and

q

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$6p+4q$	4
Notes
Since $72=2^3\cdot3^2$ , $\log_a72=3p+2q$ . Also $\log_{\sqrt a}72=\dfrac{\log_a72}{\log_a\sqrt a}=\dfrac{3p+2q}{1/2}=6p+4q$ .

(4 marks)

Tier 3 · Hard

Solve

\log_x16=\log_2x

, giving all real values of

x

allowed by the logarithms.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x=\dfrac14$ or $x=4$	5
Notes
The base requires $x>0$ and $x\neq1$ . Let $p=\log_2x$ , so $p\neq0$ . By change of base, $\log_x16=\dfrac{\log_216}{\log_2x}=4/p$ . Hence $4/p=p$ , so $p^2=4$ and $p=\pm2$ . Therefore $x=2^p$ gives $x=1/4$ or $x=4$ , both valid.

(5 marks)

Tier 2 · Standard

Solve

\ln(x+6)=2\ln x

, giving all solutions permitted by the original logarithms.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$x=3$	4
Notes
The original logarithms require $x>0$ . By the power law, $2\ln x=\ln(x^2)$ , so $x+6=x^2$ . Hence $(x-3)(x+2)=0$ , giving the candidates $x=3$ and $x=-2$ . The value $x=-2$ is outside the original domain, so the only solution is $x=3$ .

(4 marks)

Tier 3 · Hard

Positive numbers

p

and

q

satisfy

\log_2p+\log_4q=5

and

\log_8p-\log_2q=-1

. Find

p

and

q

exactly.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$p=2^{27/7}$ and $q=2^{16/7}$	5
Notes
Let $P=\log_2p$ and $Q=\log_2q$ . Change of base gives $\log_4q=Q/2$ and $\log_8p=P/3$ . The equations become $P+Q/2=5$ and $P/3-Q=-1$ . The second gives $Q=1+P/3$ . Substitution into the first gives $P+(1+P/3)/2=5$ , so $P=27/7$ and then $Q=16/7$ . Therefore $p=2^P=2^{27/7}$ and $q=2^Q=2^{16/7}$ .

(5 marks)

Tier 3 · Hard

Solve the inequality

\ln\!\left(\dfrac{x-1}{x+2}\right)>\ln\!\left(\dfrac12\right)

, giving the complete set of real values of

x

permitted by the original logarithm.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$x<-2$ or $x>4$	5
Notes
The logarithm requires $(x-1)/(x+2)>0$ , giving $x<-2$ or $x>1$ . Since $\ln x$ is increasing, the inequality is equivalent on this domain to $(x-1)/(x+2)>1/2$ . This rearranges without losing the denominator sign by writing $(x-1)/(x+2)-1/2=(x-4)/(2(x+2))>0$ . A sign analysis gives $x<-2$ or $x>4$ , both of which satisfy the original domain.

(5 marks)

Q10

Tier 3 · Hard

10.

The equation

\ln(x-1)+\ln(5-x)=k

is considered for real

x

and real

k

. Determine the values of

k

for which the equation has two distinct real solutions, exactly one real solution, or no real solutions. Find the solutions when

k=\ln3

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Two distinct solutions when $k<\ln4$ Exactly one solution when $k=\ln4$ No solutions when $k>\ln4$ When $k=\ln3$ , $x=2$ or $x=4$	6
Notes
The original logarithms require $1<x<5$ . Combining them gives $\ln((x-1)(5-x))=k$ . Now $(x-1)(5-x)=4-(x-3)^2$ , which is positive on the domain and has maximum $4$ at $x=3$ . Therefore its logarithm has range $(-\infty,\ln4]$ . Values below the maximum occur twice, the maximum once, and larger values never. If $k=\ln3$ , then $4-(x-3)^2=3$ , so $(x-3)^2=1$ and $x=2$ or $x=4$ .

(6 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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