Exponential Growth and Decay Practice Problems
Exponential growth and decay are fundamental mathematical concepts that appear in various real-world scenarios, from population biology to financial investments and radioactive decay. Because of that, understanding these patterns is crucial for solving practical problems across multiple disciplines. This complete walkthrough provides detailed practice problems with solutions to help you master exponential functions Easy to understand, harder to ignore. Simple as that..
Understanding Exponential Growth and Decay
Before diving into practice problems, let's establish the foundational formulas:
Exponential Growth: $y = a(1 + r)^t$
Exponential Decay: $y = a(1 - r)^t$
Where:
- $a$ = initial amount
- $r$ = rate of growth/decay (as a decimal)
- $t$ = time periods
- $y$ = final amount
For continuous growth/decay, we use: $A = Pe^{rt}$
Practice Problems with Solutions
Problem 1: Population Growth
A small town has a population of 5,000 people. If the population grows at a rate of 3% per year, what will the population be after 10 years?
Solution: Using the exponential growth formula: $y = a(1 + r)^t$
- $a = 5,000$
- $r = 0.03$
- $t = 10$
$y = 5,000(1 + 0.03)^{10}$ $y = 5,000(1.Which means 03)^{10}$ $y = 5,000(1. 3439)$ $y = 6,719 Worth keeping that in mind..
The population will be approximately 6,720 people after 10 years.
Problem 2: Compound Interest
Sarah invests $2,500 in a savings account that pays 4.5% interest compounded annually. How much will she have in the account after 8 years?
Solution: Using the compound interest formula: $A = P(1 + r)^t$
- $P = 2,500$
- $r = 0.045$
- $t = 8$
$A = 2,500(1 + 0.Here's the thing — 045)^8$ $A = 2,500(1. 045)^8$ $A = 2,500(1.
Sarah will have $3,556 in her account after 8 years Small thing, real impact..
Problem 3: Radioactive Decay
A sample of carbon-14 has 80 grams initially. If the half-life of carbon-14 is 5,730 years, how much remains after 11,460 years?
Solution: First, determine the decay rate. Since 11,460 years equals two half-lives: After first half-life: 80 ÷ 2 = 40 grams After second half-life: 40 ÷ 2 = 20 grams
Alternatively, using the continuous decay formula: $A = Pe^{-rt}$
To find $r$: When $t = 5,730$, $A = 40$ $40 = 80e^{-r(5,730)}$ $0.On top of that, 5 = e^{-r(5,730)}$ $\ln(0. 5) = -r(5,730)$ $r = \frac{\ln(2)}{5,730} = 0 Practical, not theoretical..
$A = 80e^{-0.000121(11,460)}$ $A = 80e^{-1.386}$ $A = 80(0.
20 grams of carbon-14 remain after 11,460 years It's one of those things that adds up..
Problem 4: Drug Concentration Decay
A patient takes a medication where 25% of the drug breaks down each hour. If the initial dose is 200 mg, how much remains after 5 hours?
Solution: Using exponential decay: $y = a(1 - r)^t$
- $a = 200$
- $r = 0.25$
- $t = 5$
$y = 200(1 - 0.Which means 25)^5$ $y = 200(0. 75)^5$ $y = 200(0.2373)$ $y = 47.
Approximately 47.5 mg of the drug remains after 5 hours.
Problem 5: Bacterial Growth
A bacterial culture starts with 500 cells and doubles every 3 hours. How many bacteria will there be after 12 hours?
Solution: First, find the growth rate. Doubling means the growth factor is 2. Number of doubling periods in 12 hours: $12 ÷ 3 = 4$
$y = a \times 2^t$ $y = 500 \times 2^4$ $y = 500 \times 16$ $y = 8,000$
There will be 8,000 bacteria after 12 hours Nothing fancy..
Problem 6: Depreciation of Vehicle Value
A new car costs $28,000 and depreciates at 15% per year. What is its value after 6 years?
Solution: Using exponential decay: $y = a(1 - r)^t$
- $a = 28,000$
- $r = 0.15$
- $t = 6$
$y = 28,000(1 - 0.In real terms, 15)^6$ $y = 28,000(0. 85)^6$ $y = 28,000(0.
The car's value will be approximately $10,567 after 6 years It's one of those things that adds up..
Problem 7: Continuous Growth - Investment
An investor puts $10,000 in a fund with continuous compounding at 6% annual interest. Find the value after 7 years using $A = Pe^{rt}$ Simple, but easy to overlook. Surprisingly effective..
Solution:
- $P = 10,000$
- $r = 0.06$
- $t = 7$
$A = 10,000e^{0.06(7)}$ $A = 10,000e^{0.42}$ $A = 10,000(1 Worth keeping that in mind. Simple as that..
The investment will be worth $15,220 after 7 years.
Problem 8: Computer Memory Growth
Computer memory capacity doubles every 18 months. If a system had 8 GB of RAM in 2015, what capacity might be expected in 2023?
Solution: Time period: 2023 - 2015 = 8 years = 96 months Number of 18-month periods: $96 ÷ 18 = 5.33$
Using the growth formula: $y = a \times 2^t$ $y = 8 \times 2^{5.33}$ $y = 8 \times 40.3$ $y = 322.
We might expect approximately 322 GB of RAM capacity Simple, but easy to overlook..
Common Types of Exponential Problems
Type 1: Population/Compound Growth
Look for phrases like "grows at a rate," "increases by," or "compounds annually." Use $y = a(1 + r)^t$.
Type 2: Decay/Radioactive Problems
Watch for "decreases," "decays," "half-life," or "depreciates."
