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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).

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Laws of logarithms

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

Explanation

  • For positive arguments, log⁡ax+log⁡ay=log⁡a(xy)\log_a x+\log_a y=\log_a(xy), log⁡ax−log⁡ay=log⁡a(x/y)\log_a x-\log_a y=\log_a(x/y) and klog⁡ax=log⁡a(xk)k\log_a x=\log_a(x^k); also log⁡aa=1\log_a a=1 and log⁡a1=0\log_a1=0.
  • To expand a logarithm, turn products into sums, quotients into differences and powers into coefficients; to condense, apply those steps in reverse.
  • A negative coefficient represents a reciprocal power, for example −12log⁡ax=log⁡a(x−1/2)=log⁡a(1/x)-\tfrac12\log_a x=\log_a(x^{-1/2})=\log_a(1/\sqrt{x}).
  • The laws combine logarithms, not their arguments: log⁡(x+y)\log(x+y) cannot be split into log⁡x+log⁡y\log x+\log y, and every original logarithm argument must remain positive.
  • Check the final form by expanding it back again.

Worked example

Expand ln⁡ ⁣(x3yz2)\ln\!\left(\dfrac{x^3\sqrt{y}}{z^2}\right), where xx, yy and zz are positive.

  1. 1.The quotient gives subtraction, the product gives addition, and powers become coefficients: ln⁡(x3)+ln⁡(y1/2)−ln⁡(z2)=3ln⁡x+12ln⁡y−2ln⁡z\ln(x^3)+\ln(y^{1/2})-\ln(z^2)=3\ln x+\tfrac12\ln y-2\ln z.

Answer: 3ln⁡x+12ln⁡y−2ln⁡z3\ln x+\dfrac12\ln y-2\ln z

Common mistakes

  • Don't split log⁡(x+y)\log(x+y) into log⁡x+log⁡y\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.

Exam tip

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

Worked practice

Q1
Tier 1 · Easy

1.

Write ln⁡12+ln⁡3−ln⁡2\ln12+\ln3-\ln2 as a single logarithm.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • ln⁡18\ln18
2
Notes
Use product and quotient laws: ln⁡12+ln⁡3−ln⁡2=ln⁡(12×3/2)=ln⁡18\ln12+\ln3-\ln2=\ln(12\times3/2)=\ln18.

(2 marks)

Q2
Tier 2 · Standard

2.

Solve 2ln⁡(x+1)−ln⁡x=ln⁡82\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
QuestionSchemeMarks
2
  • x=3−22x=3-2\sqrt2 or x=3+22x=3+2\sqrt2
5
Notes
The logarithms require x>0x>0. Combine them to get ln⁡ ⁣((x+1)2/x)=ln⁡8\ln\!\left((x+1)^2/x\right)=\ln8, so (x+1)2=8x(x+1)^2=8x. Hence x2−6x+1=0x^2-6x+1=0, giving x=3±22x=3\pm2\sqrt2. Both values are positive, so both satisfy the domain and are valid.

(5 marks)

Q3
Tier 3 · Hard

3.

Solve log⁡3(x−1)+log⁡3(x+3)=2\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
QuestionSchemeMarks
3
  • x=−1+13x=-1+\sqrt{13}
5
Notes
The original arguments require x>1x>1. Combine the logarithms: log⁡3((x−1)(x+3))=2\log_3((x-1)(x+3))=2, so (x−1)(x+3)=9(x-1)(x+3)=9. Hence x2+2x−12=0x^2+2x-12=0, giving x=−1±13x=-1\pm\sqrt{13}. Only −1+13>1-1+\sqrt{13}>1, so the other algebraic root is rejected.

(5 marks)

Q4
Tier 1 · Easy

4.

Given that p>0p>0 and q>0q>0, express 3log⁡ap−12log⁡aq3\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
QuestionSchemeMarks
4
  • log⁡a ⁣(p3q)\log_a\!\left(\dfrac{p^3}{\sqrt q}\right)
2
Notes
Use the power law first: 3log⁡ap=log⁡a(p3)3\log_a p=\log_a(p^3) and 12log⁡aq=log⁡a(q)\tfrac12\log_a q=\log_a(\sqrt q). The subtraction law then gives log⁡a(p3/q)\log_a(p^3/\sqrt q).

(2 marks)

Q5
Tier 2 · Standard

5.

Given that log⁡a2=p\log_a2=p and log⁡a3=q\log_a3=q, express log⁡a72\log_{\sqrt a}72 in terms of pp and qq.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • 6p+4q6p+4q
4
Notes
Since 72=23⋅3272=2^3\cdot3^2, log⁡a72=3p+2q\log_a72=3p+2q. Also log⁡a72=log⁡a72log⁡aa=3p+2q1/2=6p+4q\log_{\sqrt a}72=\dfrac{\log_a72}{\log_a\sqrt a}=\dfrac{3p+2q}{1/2}=6p+4q.

(4 marks)

Q6
Tier 3 · Hard

6.

Solve log⁡x16=log⁡2x\log_x16=\log_2x, giving all real values of xx allowed by the logarithms.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • x=14x=\dfrac14 or x=4x=4
5
Notes
The base requires x>0x>0 and x≠1x\neq1. Let p=log⁡2xp=\log_2x, so p≠0p\neq0. By change of base, log⁡x16=log⁡216log⁡2x=4/p\log_x16=\dfrac{\log_216}{\log_2x}=4/p. Hence 4/p=p4/p=p, so p2=4p^2=4 and p=±2p=\pm2. Therefore x=2px=2^p gives x=1/4x=1/4 or x=4x=4, both valid.

(5 marks)

Q7
Tier 2 · Standard

7.

Solve ln⁡(x+6)=2ln⁡x\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
QuestionSchemeMarks
7
  • x=3x=3
4
Notes
The original logarithms require x>0x>0. By the power law, 2ln⁡x=ln⁡(x2)2\ln x=\ln(x^2), so x+6=x2x+6=x^2. Hence (x−3)(x+2)=0(x-3)(x+2)=0, giving the candidates x=3x=3 and x=−2x=-2. The value x=−2x=-2 is outside the original domain, so the only solution is x=3x=3.

(4 marks)

Q8
Tier 3 · Hard

8.

Positive numbers pp and qq satisfy log⁡2p+log⁡4q=5\log_2p+\log_4q=5 and log⁡8p−log⁡2q=−1\log_8p-\log_2q=-1. Find pp and qq exactly.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • p=227/7p=2^{27/7} and q=216/7q=2^{16/7}
5
Notes
Let P=log⁡2pP=\log_2p and Q=log⁡2qQ=\log_2q. Change of base gives log⁡4q=Q/2\log_4q=Q/2 and log⁡8p=P/3\log_8p=P/3. The equations become P+Q/2=5P+Q/2=5 and P/3−Q=−1P/3-Q=-1. The second gives Q=1+P/3Q=1+P/3. Substitution into the first gives P+(1+P/3)/2=5P+(1+P/3)/2=5, so P=27/7P=27/7 and then Q=16/7Q=16/7. Therefore p=2P=227/7p=2^P=2^{27/7} and q=2Q=216/7q=2^Q=2^{16/7}.

(5 marks)

Q9
Tier 3 · Hard

9.

Solve the inequality ln⁡ ⁣(x−1x+2)>ln⁡ ⁣(12)\ln\!\left(\dfrac{x-1}{x+2}\right)>\ln\!\left(\dfrac12\right), giving the complete set of real values of xx permitted by the original logarithm.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • x<−2x<-2 or x>4x>4
5
Notes
The logarithm requires (x−1)/(x+2)>0(x-1)/(x+2)>0, giving x<−2x<-2 or x>1x>1. Since ln⁡x\ln x is increasing, the inequality is equivalent on this domain to (x−1)/(x+2)>1/2(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(x-1)/(x+2)-1/2=(x-4)/(2(x+2))>0. A sign analysis gives x<−2x<-2 or x>4x>4, both of which satisfy the original domain.

(5 marks)

Q10
Tier 3 · Hard

10.

The equation ln⁡(x−1)+ln⁡(5−x)=k\ln(x-1)+\ln(5-x)=k is considered for real xx and real kk. Determine the values of kk for which the equation has two distinct real solutions, exactly one real solution, or no real solutions. Find the solutions when k=ln⁡3k=\ln3.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • Two distinct solutions when k<ln⁡4k<\ln4
  • Exactly one solution when k=ln⁡4k=\ln4
  • No solutions when k>ln⁡4k>\ln4
  • When k=ln⁡3k=\ln3, x=2x=2 or x=4x=4
6
Notes
The original logarithms require 1<x<51<x<5. Combining them gives ln⁡((x−1)(5−x))=k\ln((x-1)(5-x))=k. Now (x−1)(5−x)=4−(x−3)2(x-1)(5-x)=4-(x-3)^2, which is positive on the domain and has maximum 44 at x=3x=3. Therefore its logarithm has range (−∞,ln⁡4](-\infty,\ln4]. Values below the maximum occur twice, the maximum once, and larger values never. If k=ln⁡3k=\ln3, then 4−(x−3)2=34-(x-3)^2=3, so (x−3)2=1(x-3)^2=1 and x=2x=2 or x=4x=4.

(6 marks)

Verified exam appearances

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Other points in 6 Exponentials and logarithms

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