Exponential & Logarithmic Models — Challenge

Growth and decay that depend on how much is already there, and the logs that solve them.

What these problems cover

These questions use 5 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Problem 1

If y=3· 2x, what is y when x=0?

  1. 0
  2. 1
  3. 2
  4. 3
  5. 6
Worked answer

3

Any non-zero number to the power 0 is 1, so 20 = 1.

Then y = 3· 1 = 3.

In general the starting value of y = a· bx is a.

Problem 2

A car worth $20,000 loses 10% of its value each year. What is it worth in dollars after one year?

  1. $17,000
  2. $18,000
  3. $19,000
  4. $19,800
  5. $22,000
Worked answer

$18,000

Losing 10% leaves 90% of the value, so multiply by 0.90.

20,000· 0.90 = 18,000 dollars.

Problem 3

A population of 200 triples every decade. What is it after two decades?

  1. 400
  2. 600
  3. 1200
  4. 1800
  5. 3600
Worked answer

1800

Tripling twice multiplies by 32 = 9.

200· 9 = 1800.

Multiplying by 3 and then adding another 3 times the start would give 1200, which treats the growth as additive.

Problem 4

An account holds $500 and earns 5% interest a year, compounded annually. What is the balance in dollars after two years?

  1. $525.00
  2. $550.00
  3. $551.25
  4. $552.50
  5. $605.00
Worked answer

$551.25

Each year multiplies the balance by 1.05, so two years multiply by 1.052 = 1.1025.

500· 1.1025 = 551.25 dollars.

Simple interest would give 550, missing the 1.25 of interest earned on the first year's interest.

Problem 5

A quantity halves every 3 hours. What fraction of it remains after 12 hours?

  1. 14
  2. 18
  3. 116
  4. 112
  5. 132
Worked answer

116

Twelve hours contains 123 = 4 halving periods.

Halving four times leaves (12)4 = 116.

Dividing by 4 to get 14 would treat the decay as linear.

Problem 6

The model y=100(1.2)x describes growth at what rate per step?

  1. 20%
  2. 1.2%
  3. 120%
  4. 80%
  5. 12%
Worked answer

20%

A base of 1.2 means each step multiplies the quantity by 1.2.

Multiplying by 1.2 is the same as keeping 100% and adding another 20%.

So the growth rate is 20% per step. A base of 0.8 would instead mean a 20% decline.

Problem 7

A bacteria culture doubles every 20 minutes. If there are 400 cells now, how many will there be in 2 hours?

  1. 2400
  2. 4800
  3. 12800
  4. 25600
  5. 51200
Worked answer

25600

Two hours is 120 minutes, which is 120 ÷ 20 = 6 doublings. Six doublings multiply by 26=64. 400 × 64 = 25600. Adding 400 six times would give only 2800 — doubling is not repeated addition.

Problem 8

A substance decays to 18 of its original amount in 12 years. What is its half-life?

  1. 2 years
  2. 3 years
  3. 4 years
  4. 6 years
  5. 8 years
Worked answer

4 years

18=(12)3, so three half-lives have passed. Three half-lives take 12 years. So one half-life is 12 ÷ 3 = 4 years.

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