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Exponential growth and decay word problems: build y = a · bt, then solve for t

Population, money, radioactive samples, and the value of a used car all follow y = a · bt. The points get lost when the rate is used as the base, when decay is written with the wrong factor, when the log is taken before the starting amount is divided off, and when the half-life exponent is t instead of t/h.

Short answer

Start with y = a · bt, where a is the starting amount and b is the factor per time unit. Growth of r percent per year means b = 1 + r; decay of r percent means b = 1 − r. A half-life of h years means y = a(12)t/h. To find y, plug in t. To find t, divide by a first, then take a log of both sides and bring the exponent down.

Why it works

"Grows 4% per year" means each year's amount is 104% of the previous one: multiply by 1.04. After t years you've multiplied by 1.04 t times, which is 1.04t. Decay is the same with a factor below 1: losing 15% leaves 85%, so multiply by 0.85 each year. The percent is never the base by itself; the base is what's left or what it becomes.

Solving for t is a one-step exponential equation once the power is alone: bt = c gives t = ln cln b, by the power rule for logarithms. Any log works as long as you use the same one on both sides.

Example 1: percent growth, find the amount

A town of 12,000 grows 4% per year. What is the population after 6 years?

  1. P = 12000(1.04)ta = 12,000, b = 1 + 0.04 = 1.04.
  2. P = 12000(1.04)6t = 6.
  3. P ≈ 12000(1.265319) ≈ 15,184Compute the power first, then multiply. Rounding to a whole number of people at the end.

Example 2: percent decay, find the time

A car bought for $28,000 loses 15% of its value each year. When is it worth $10,000?

  1. 28000(0.85)t = 10000Decay factor 1 − 0.15 = 0.85. The 10,000 is the target value, so it goes on the other side.
  2. 0.85t = 1000028000 ≈ 0.357143Divide by the starting amount before taking any log.
  3. t · ln 0.85 = ln 0.357143Take ln of both sides; the power rule brings t down in front.
  4. t = ln 0.357143ln 0.85 ≈ −1.02962−0.162519 ≈ 6.34 yearsBoth logs are negative, so t is positive. Check: 28000(0.85)6.34 ≈ 9,992, which is 10,000 to the rounding.

Example 3: half-life

A 200 mg sample of cobalt-60 has a half-life of 5.3 years. How much is left after 12 years, and how long until 25 mg is left?

  1. A = 200(12)t/5.3The exponent counts how many half-lives have passed: t years ÷ 5.3 years per half-life.
  2. A = 200(12)12/5.3 ≈ 200(0.5)2.2642 ≈ 200(0.2082) ≈ 41.6 mg12 years is about 2.26 half-lives. Two half-lives would leave 50 mg; a bit more than two leaves a bit less. Reasonable.
  3. 200(12)t/5.3 = 25, so (12)t/5.3 = 18Divide by 200 first. 25/200 = 1/8.
  4. t5.3 = 3, so t = 15.9 years1/8 = (1/2)³, so no log is needed: exactly three half-lives. If the ratio weren't a nice power of ½, take ln of both sides and divide by ln(½).

Example 4: find the factor from two data points

A bacteria culture grows from 500 to 2,000 cells in 3 hours. What is the hourly growth rate, and how many cells are there after 6 hours?

  1. 500 · b3 = 2000a = 500, t = 3, y = 2000. The unknown is the base.
  2. b3 = 4, so b = 41/3 ≈ 1.5874Divide by 500, then take the cube root (a 1/3 power). Keep the exact form 41/3 for later steps.
  3. Rate ≈ 58.7% per hourb − 1 = 0.5874. The rate is the part above 1.
  4. y = 500 · (41/3)6 = 500 · 42 = 8,000Six hours is two three-hour periods, and each one quadruples. Using the exact base makes this come out clean.

Common mistakes, and the exact line they happen on

1. Using the rate as the base

Set up the model in Example 1.

  1. P = 12000(0.04)tThis is the mistake. 0.046 is about 0.000004, so this says the town has shrunk to a fraction of one person. Multiplying by 0.04 keeps 4% of the town each year; the problem says the town gains 4%.
  2. P = 12000(1.04)t100% of what was there plus 4% more.

Sanity check every base: growth needs b > 1, decay needs 0 < b < 1. A base of 0.04 for growth, or 1.15 for a car losing value, is wrong before any arithmetic.

2. Writing decay with the wrong factor

Set up the model in Example 2.

  1. V = 28000(0.15)tThis is the mistake. After one year this gives $4,200, an 85% loss. Losing 15% means keeping 85%.
  2. V = 28000(1.15)tThe other version. This car gains value. Subtracting the rate was skipped.
  3. V = 28000(0.85)t1 − 0.15.

3. Taking the log before dividing by a

Solve 28000(0.85)t = 10000 (Example 2).

  1. t · ln(28000 · 0.85) = ln 10000This is the mistake. The power rule moves an exponent that applies to the whole thing inside the log. Here the exponent applies only to 0.85, not to 28000 · 0.85. The correct expansion of the left side is ln 28000 + t · ln 0.85, which is longer, not shorter.
  2. 0.85t = 1000028000, then t · ln 0.85 = ln(1000028000)Divide first so the power is alone. Then the exponent really does apply to everything inside the log.

4. Using t instead of t/h in a half-life problem

Find the amount after 12 years in Example 3.

  1. A = 200(12)12 ≈ 0.049 mgThis is the mistake. This halves the sample 12 times, as if the half-life were one year. The half-life is 5.3 years, so 12 years is 12 ÷ 5.3 ≈ 2.26 halvings.
  2. A = 200(12)12/5.3 ≈ 41.6 mgThe exponent is the number of half-lives, not the number of years.

5. Rounding the base too early

Find the 6-hour count in Example 4.

  1. b ≈ 1.59, so y = 500(1.59)6 ≈ 8,079This is the mistake. The exact answer is 8,000. Rounding b to two decimals and then raising it to the sixth power magnifies the rounding error by a lot, and the error grows with t.
  2. y = 500 · (41/3)6 = 500 · 42 = 8,000Carry the base as 41/3, or at least to four or five decimals, and round only the final answer.

Practice

  1. An investment of $5,000 grows 8% per year. What is it worth after 10 years?

    Show answer

    5000(1.08)10 ≈ $10,795.

  2. How long does it take $1,200 at 6% per year, compounded annually, to reach $2,000?

    Show answer

    1.06t = 2000/1200 = 5/3, so t = ln(5/3) ÷ ln 1.06 ≈ 8.77 years. If interest is only credited at the end of each year, it crosses $2,000 at the end of year 9.

  3. Caffeine has a half-life of about 5 hours. How much of a 160 mg coffee is left after 8 hours?

    Show answer

    160(½)8/5 ≈ 160(0.3299) ≈ 52.8 mg.

  4. A town shrinks from 40,000 to 34,000 in 5 years. What is the annual percent decay?

    Show answer

    b5 = 34000/40000 = 0.85, so b = 0.851/5 ≈ 0.968, a decay of about 3.2% per year.

  5. At 7% growth per year, how long does an amount take to double?

    Show answer

    1.07t = 2, so t = ln 2 ÷ ln 1.07 ≈ 10.24 years. The starting amount cancels, so it doesn't matter what it was.

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