Exponential Function Word Problems with Answers
Exponential functions appear in many real‑world situations, from population growth and radioactive decay to compound interest and the spread of viruses. In real terms, understanding how to translate a word problem into an exponential model, solve for the unknown, and interpret the result is a key skill in algebra and pre‑calculus. Below you will find a step‑by‑step guide, several worked examples, and a set of practice problems with detailed answers to reinforce your learning.
How to Approach Exponential Function Word Problems
- Read the problem carefully – Identify what quantity is changing exponentially and what information is given (initial value, growth/decay rate, time period).
- Choose the appropriate form –
- For growth: (y = a \cdot b^{t}) where (b > 1).
- For decay: (y = a \cdot b^{t}) where (0 < b < 1).
- Often it is convenient to write the rate as a percent: (b = 1 + r) for growth or (b = 1 - r) for decay, with (r) expressed as a decimal.
- Assign variables – Let (y) be the amount after time (t), (a) the initial amount, and (t) the time in the units given (years, months, hours, etc.).
- Set up the equation – Plug the known values into the exponential formula.
- Solve for the unknown – Use logarithms when the variable is in the exponent:
[ t = \frac{\ln\left(\frac{y}{a}\right)}{\ln(b)} ]
or solve for (y) directly if the exponent is known. - Check the answer – Verify that the solution makes sense in the context (e.g., a population cannot be negative, a decaying substance approaches zero).
Worked Examples
Example 1: Population Growth
Problem: A small town has a population of 12,000 people. The population grows at a rate of 3.5% per year. What will the population be after 8 years?
Solution:
- Identify the initial amount (a = 12{,}000).
- Growth rate (r = 3.5% = 0.035).
- Growth factor (b = 1 + r = 1.035).
- Time (t = 8) years.
Plug into the formula:
[ y = 12{,}000 \times (1.035)^{8} ]
Calculate:
[ (1.035)^{8} \approx 1.316 ]
[ y \approx 12{,}000 \times 1.316 = 15{,}792 ]
Answer: Approximately 15,792 people will live in the town after 8 years And it works..
Example 2: Radioactive Decay
Problem: A sample of a radioactive isotope has an initial mass of 200 grams. Its half‑life is 5 years. How much of the isotope remains after 15 years?
Solution:
Half‑life means that after each 5‑year interval the mass is halved. This can be expressed with a decay factor (b = \left(\frac{1}{2}\right)^{\frac{t}{5}}) Practical, not theoretical..
Alternatively, use the exponential decay form (y = a \cdot b^{t}) where (b = \left(\frac{1}{2}\right)^{1/5}).
Compute (b):
[ b = \left(\frac{1}{2}\right)^{0.2} \approx 0.87055 ]
Now apply the formula with (a = 200) g and (t = 15) years:
[ y = 200 \times (0.87055)^{15} ]
[ (0.87055)^{15} \approx 0.125 ]
[ y \approx 200 \times 0.125 = 25 \text{ grams} ]
Answer: About 25 grams of the isotope remain after 15 years (which is exactly three half‑lives, so (200 \times (1/2)^3 = 25) g).
Example 3: Compound Interest
Problem: You invest $4,500 in an account that offers 6% annual interest compounded monthly. How much money will be in the account after 4 years?
Solution:
For compound interest the formula is
[ A = P \left(1 + \frac{r}{n}\right)^{nt} ]
where
- (P = 4{,}500) (principal)
- (r = 0.06) (annual rate)
- (n = 12) (compounding periods per year)
- (t = 4) years
Plug in:
[ A = 4{,}500 \left(1 + \frac{0.06}{12}\right)^{12 \times 4} = 4{,}500 \left(1 + 0.005\right)^{48} = 4{,}500 \times (1 Practical, not theoretical..
Calculate ((1.005)^{48} \approx 1.27024).
[ A \approx 4{,}500 \times 1.27024 = 5{,}716.08 ]
Answer: The account will contain approximately $5,716.08 after 4 years Most people skip this — try not to. Nothing fancy..
Example 4: Finding Time from Growth
Problem: A bacterial culture starts with 500 cells and doubles every 20 minutes. How long will it take for the culture to reach 80,000 cells?
Solution:
Doubling every 20 minutes means the growth factor per 20‑minute interval is 2. Let (n) be the number of 20‑minute intervals needed That alone is useful..
[ 500 \times 2^{n} = 80{,}000 ]
Divide both sides by 500:
[ 2^{n} = \frac{80{,}000}{500} = 160 ]
Take log base 2 of both sides:
[ n = \log_{2}(160) = \frac{\ln(160)}{\ln(2)} \approx \frac{5.Think about it: 075}{0. 693} \approx 7 And it works..
So about 7.32 intervals are required
