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Exponential Functions: Graphs, Equations and Models — Answers

Answer sheet with worked solutions for the Year 12 exponentials & logarithms worksheet.

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Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Exponentials & Logarithms

Exponential Functions: Graphs, Equations and Models — Answers

Part A: Graphs of exponential functions

For y = bx with b > 0 and b ≠ 1: the domain is all real x, the range is y > 0, the y-intercept is (0, 1) and the x-axis (y = 0) is a horizontal asymptote. If b > 1 the graph increases (growth); if 0 < b < 1 it decreases (decay).

Transformations work as for other functions: y = bx + c moves the graph and its asymptote up c units; y = a × bx stretches it vertically; y = b−x reflects it in the y-axis.

Worked example. y = 2x + 3 has y-intercept 20 + 3 = 4, asymptote y = 3 and range y > 3.

1.The graph of y = 3x − 5 has y-intercept (answer space) and horizontal asymptote y = (answer space).

Answer: −4, −5

At x = 0: y = 30 − 5 = 1 − 5 = −4. The graph of y = 3x has asymptote y = 0; shifting down 5 moves it to y = −5.

2.Which statement describes the graph of y = (0.5)x?
  1. ADecreasing, passing through (0, 1)
  2. BIncreasing, passing through (0, 1)
  3. CDecreasing, passing through (0, 0.5)
  4. DIncreasing, passing through (1, 0)

Answer: Decreasing, passing through (0, 1)

The base 0.5 is between 0 and 1, so the function decreases (decay). Every graph y = bx passes through (0, 1) because b0 = 1.

3.The graph of y = 4−x is the reflection of the graph of y = 4x in the y-axis.

Answer: True

Replacing x with −x reflects a graph in the y-axis. Note that 4−x = ()x, a decreasing function.

4.State the range of y = 2 × 5x + 1.

Answer: y > 1

  1. 5x > 0 for all x, so 2 × 5x > 0.
  2. Adding 1: y > 1 (the asymptote y = 1 is never reached)

Part B: Solving exponential equations

Same base: if both sides can be written as powers of the same number, equate the indices: bm = bn means m = n.

Different bases: take logarithms (ln or log) of both sides and use log(bx) = x log b. So bx = c gives x = [frac ln c/ln b]. Get the power on its own first.

Quadratic form: b2x = (bx)2, so an equation like b2x + pbx + q = 0 becomes a quadratic in u = bx. Remember that bx > 0, so reject negative values of u.

Worked example. Solve 9x + 1 = 27x. Powers of 3: 32x + 2 = 33x, so 2x + 2 = 3x and x = 2.

5.Solve 23x − 1 = 32.

Answer: 2

  1. 32 = 25, so 3x − 1 = 5
  2. 3x = 6, x = 2
6.Solve 4x = 8x − 1.

Answer: 3

  1. Write both sides as powers of 2: 22x = 23(x − 1) = 23x − 3
  2. 2x = 3x − 3, so x = 3
7.Solve 5x = 40, correct to 2 decimal places.

Answer: 2.29

  1. Take logarithms of both sides: x ln 5 = ln 40
  2. x = = = 2.2920…
  3. x ≈ 2.29
8.Solve 3 × 2x = 51, correct to 2 decimal places.

Answer: 4.09

  1. Divide by 3 first: 2x = 17
  2. x = = 4.0874…
  3. x ≈ 4.09
9.Solve 72x = 3x + 1, correct to 3 decimal places.

Answer: 0.393

  1. Take logarithms: 2x ln 7 = (x + 1) ln 3
  2. 2x ln 7 − x ln 3 = ln 3, so x(2 ln 7 − ln 3) = ln 3
  3. x = = = 0.39332…
  4. x ≈ 0.393
10.Solve 22x − 6 × 2x + 8 = 0.

Answer: 1, 2

  1. Let u = 2x, so 22x = u2: u2 − 6u + 8 = 0
  2. (u − 2)(u − 4) = 0, so u = 2 or u = 4
  3. 2x = 2 gives x = 1; 2x = 4 gives x = 2
  4. x = 1, 2

Part C: Exponential growth and decay models

A quantity that changes by the same percentage each period is modelled by y = a × b[sup t], where a is the starting amount. Growth of r% per period gives b = 1 + ; decay of r% gives b = 1 − .

If a quantity doubles every d units of time, y = a × 2t/d. If it has a half-life of h, y = a × ()t/h.

Worked example. A car bought for $30 000 loses 15% of its value each year. After 4 years V = 30 000 × 0.854 = $15 660.19.

11.A culture starts with 500 bacteria and doubles every 3 hours, so N = 500 × 2t/3 after t hours. Find the number of bacteria after 10 hours, to the nearest whole number.

Answer: 5040

  1. N = 500 × 210/3 = 500 × 10.0793…
  2. = 5039.68…, so about 5040 bacteria
12.A town of 2400 people grows by 3.5% each year. After how many whole years does the population first exceed 4000?

Answer: 15

  1. P = 2400 × 1.035n. Solve 2400 × 1.035n = 4000: 1.035n =
  2. n = = 14.85…
  3. Check: after 14 years P ≈ 3885; after 15 years P ≈ 4021.
  4. The population first exceeds 4000 after 15 years.
13.A substance has a half-life of 8 days. Starting with 200 mg, how much remains after 20 days, to 1 decimal place?

Answer: 35.4 mg

  1. M = 200 × ()t/8
  2. M = 200 × 0.52.5 = 200 × 0.17677…
  3. = 35.4 mg
14.A model y = a × bx (with b > 0) passes through (1, 6) and (3, 54). Then a = (answer space) and b = (answer space).

Answer: 2, 3

ab = 6 and ab3 = 54. Dividing: b2 = 9, so b = 3 (b > 0), and a = 6 ÷ 3 = 2. So y = 2 × 3x.

15.A quantity starts at 100 and decreases by 20% every hour. Which model gives the amount Q after t hours?
  1. AQ = 100(0.8)t
  2. BQ = 100(0.2)t
  3. CQ = 100(1.2)t
  4. DQ = 100 − 0.2t

Answer: Q = 100(0.8)t

Losing 20% leaves 80% each hour, so multiply by 0.8 each hour: Q = 100(0.8)t. The model 100(0.2)t would keep only 20% each hour.

16.A quantity that grows by 10% each year doubles in exactly 10 years.

Answer: False

1.110 ≈ 2.594, so after 10 years it is about 2.6 times as large. It doubles when 1.1n = 2, that is after n = ≈ 7.27 years. Growth compounds, so it is faster than 10 lots of 10%.

Part D: Extended problem

A patient is given a 400 mg dose of a medicine. Each hour, the amount in the body falls by 15%, so after t hours the amount is A(t) = 400 × 0.85t mg.

(a) Find the amount remaining after 3 hours.

(b) Find the half-life of the medicine in the body, correct to 2 decimal places.

(c) How long, correct to 1 decimal place, until less than 50 mg remains?

(d) A second 400 mg dose is given exactly 6 hours after the first. Find the amount in the body just after the second dose, to the nearest 0.01 mg.

(e) Explain why, under this model, the first dose never disappears completely.

17.Answer parts (a) to (e), showing full working.[10 marks]

Model answer: (a) A(3) = 400 × 0.853 = 400 × 0.614125 = 245.65 mg. (b) Solve 0.85t = 0.5: t = = 4.27 hours. (c) Solve 400 × 0.85t = 50: 0.85t = 0.125, t = = 12.79…, so about 12.8 hours. (d) Just before the second dose, A(6) = 400 × 0.856 = 150.86 mg, so just after it the amount is 150.86 + 400 = 550.86 mg. (e) 0.85t > 0 for every t, so the amount gets closer and closer to 0 but never reaches it: the t-axis is a horizontal asymptote. In practice the amount becomes too small to matter.

Success criteria:

  • (a) 245.65 mg (1 mark)
  • (b) uses logarithms to find the half-life ≈ 4.27 h (2 marks)
  • (c) solves 0.85t = 0.125 to get ≈ 12.8 h (2 marks)
  • (d) finds 150.86 mg remaining and adds the new dose to get 550.86 mg (3 marks)
  • (e) explains the horizontal asymptote (2 marks)
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