A-level Mathematics

6.1 The functions a^x and e^x — Practise

Work through these on your own. Before each question, decide whether the base lies above 11 or between 00 and 11, and note whether the question asks for an exact value or a decimal.

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

  1. Work out 25×2−22^5 \times 2^{-2}.
  2. The curve y=7xy = 7^x crosses the yy-axis at one point. Write down the yy-coordinate of that point.
  3. Work out (12)−3\left(\dfrac{1}{2}\right)^{-3}.
  4. Work out 163/416^{3/4}.
  5. Use e≈2.71828e \approx 2.71828 to write e2e^2 to three significant figures.
  6. Show that 2x=exln⁡22^x = e^{x\ln 2}.
  7. Work out 343^4 and 434^3, and state which is larger.
  8. Simplify 2x+3÷2x−12^{x+3} \div 2^{x-1}.
  9. For y=(0.4)xy = (0.4)^x, work out yy when x=2x = 2 and when x=−1x = -1.
  10. The curve y=3xy = 3^x meets the line y=81y = 81 once. Find the xx-coordinate of the meeting point.
  11. Work out e0+e1e^0 + e^1 to three significant figures.
Answers and concise solutions

1. The value

Answer: Add the indices: 25×2−2=25−2=23=82^5 \times 2^{-2} = 2^{5-2} = 2^3 = 8.

2. The yy-coordinate

Answer: Every curve y=axy = a^x passes through (0,1)(0, 1), because a0=1a^0 = 1 for every positive base aa.

3. The value

Answer: A negative index turns the fraction over: (12)−3=23=8\left(\dfrac{1}{2}\right)^{-3} = 2^3 = 8.

4. The value

Answer: The power 34\dfrac{3}{4} means the fourth root, then the cube: 161/4=216^{1/4} = 2, and 23=82^3 = 8.

5. The value

Answer: e2=e×e≈7.389056e^2 = e \times e \approx 7.389056, which is 7.397.39 to three significant figures.

6. The difference

Answer: Write the base 22 as a power of ee: 2=eln⁡22 = e^{\ln 2}. Then 2x=(eln⁡2)x=exln⁡22^x = \left(e^{\ln 2}\right)^x = e^{x\ln 2}, using the index law (am)n=amn\left(a^m\right)^n = a^{mn}.

7. The two values

Answer: 34=813^4 = 81 and 43=644^3 = 64. So 343^4 is larger. A larger base does not always win: the exponent matters too.

8. The value

Answer: Subtract the indices: 2(x+3)−(x−1)=24=162^{(x+3) - (x-1)} = 2^4 = 16. The xx cancels, so the answer is a number.

9. The two values

Answer: (0.4)2=0.16=425(0.4)^2 = 0.16 = \dfrac{4}{25} and (0.4)−1=10.4=2.5=52(0.4)^{-1} = \dfrac{1}{0.4} = 2.5 = \dfrac{5}{2}. The values fall as xx grows, because the base is between 00 and 11.

10. The value of xx

Answer: Write 8181 as a power of 33: 81=3481 = 3^4. So 3x=343^x = 3^4 and x=4x = 4.

11. The value

Answer: e0=1e^0 = 1 and e1=e≈2.71828e^1 = e \approx 2.71828. Their sum is 3.718283.71828, which is 3.723.72 to three significant figures.

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