A-level Mathematics

6.4 Solving equations of the form a^x = b — Practise

Work through these on your own. Isolate the exponential before you take logarithms, and check that any answer to a logarithmic equation lies in the domain.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

  1. Solve 2x=72^x = 7, giving xx exactly.
  2. Solve 3x=203^x = 20, giving xx exactly.
  3. Solve 5x=0.15^x = 0.1, giving xx exactly.
  4. Solve ex=5e^x = 5, giving xx exactly.
  5. Solve e2x=9e^{2x} = 9, giving xx exactly.
  6. Solve e2x−5ex+6=0e^{2x} - 5e^x + 6 = 0, giving each value of xx exactly.
  7. Solve 22x−5(2x)+4=02^{2x} - 5\left(2^x\right) + 4 = 0, giving each value of xx exactly.
  8. Solve 52x−1=305^{2x-1} = 30, giving xx exactly.
  9. Solve 3x+1=1003^{x+1} = 100, giving xx exactly.
  10. Solve 2x=72^x = 7, giving xx to three significant figures.
  11. Solve ex=−1e^x = -1 for real xx, or explain why there is no solution.
  12. Solve 4x=84^x = 8 by writing both sides as powers of 22.
Answers and concise solutions

1. The value of xx

Answer: Take logarithms of both sides: xln⁡2=ln⁡7x\ln 2 = \ln 7, so x=ln⁡7ln⁡2x = \dfrac{\ln 7}{\ln 2}. This is about 2.8072.807.

2. The value of xx

Answer: Take logarithms: x=ln⁡20ln⁡3x = \dfrac{\ln 20}{\ln 3}, which is about 2.7272.727. Any base of logarithm works; the natural logarithm is the usual choice.

3. The value of xx

Answer: Since 0.1=10−10.1 = 10^{-1}, taking logarithms gives xln⁡5=−ln⁡10x\ln 5 = -\ln 10, so x=−ln⁡10ln⁡5≈−1.431x = -\dfrac{\ln 10}{\ln 5} \approx -1.431. The answer is negative because the number on the right is smaller than 11.

4. The value of xx

Answer: The natural logarithm undoes the exponential, so x=ln⁡5≈1.609x = \ln 5 \approx 1.609.

5. The value of xx

Answer: Take logarithms: 2x=ln⁡92x = \ln 9, so x=ln⁡92=ln⁡3≈1.099x = \dfrac{\ln 9}{2} = \ln 3 \approx 1.099. Halving is the last step, after the logarithm.

6. The values of xx

Answer: This is a quadratic in exe^x. Factorise: (ex−2)(ex−3)=0\left(e^x - 2\right)\left(e^x - 3\right) = 0, so ex=2e^x = 2 or ex=3e^x = 3. Each gives a logarithm: x=ln⁡2x = \ln 2 or x=ln⁡3x = \ln 3.

7. The values of xx

Answer: Write 22x2^{2x} as (2x)2\left(2^x\right)^2 and treat 2x2^x as the unknown. Factorise: (2x−1)(2x−4)=0\left(2^x - 1\right)\left(2^x - 4\right) = 0. So 2x=12^x = 1, giving x=0x = 0, or 2x=42^x = 4, giving x=2x = 2.

8. The value of xx

Answer: Take logarithms: (2x−1)ln⁡5=ln⁡30(2x - 1)\ln 5 = \ln 30, so 2x−1=ln⁡30ln⁡52x - 1 = \dfrac{\ln 30}{\ln 5} and x=1+ln⁡62ln⁡5x = 1 + \dfrac{\ln 6}{2\ln 5}. This is about 1.5571.557.

9. The value of xx

Answer: Take logarithms: (x+1)ln⁡3=ln⁡100(x + 1)\ln 3 = \ln 100, so x=ln⁡100ln⁡3−1≈3.192x = \dfrac{\ln 100}{\ln 3} - 1 \approx 3.192. The −1-1 comes from the exponent x+1x + 1.

10. The value of xx

Answer: x=ln⁡7ln⁡2=2.80735…x = \dfrac{\ln 7}{\ln 2} = 2.80735\ldots, which is 2.812.81 to three significant figures.

11. The solution set

Answer: The graph of y=exy = e^x lies above the xx-axis for every real xx, so it never equals −1-1. The solution set is empty. This is why a logarithmic equation is checked: only a positive input is allowed.

12. The value of xx

Answer: Write 4=224 = 2^2 and 8=238 = 2^3. Then (22)x=23\left(2^2\right)^x = 2^3, so 22x=232^{2x} = 2^3 and 2x=32x = 3. The solution is x=32=1.5x = \dfrac{3}{2} = 1.5. The computer also reports this as ln⁡2ln⁡4+1\dfrac{\ln 2}{\ln 4} + 1, which is the same number.

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