Exponential Function Word Problems

Show your work for each problem.

1.
A phone loses 15% of its value every year. Which factor $b$ do you use in the model $V(t) = a \cdot b^t$ ?
1. a)0.15
2. b)1.15
3. c)0.85
4. d)85
Answer: 0.85
Decay: $b = 1 - r = 1 - 0.15 = 0.85$ . Each year 85% of the value stays. $0.15$ is the rate, not the factor, and $1.15$ would mean growth.
2.
The number of fish in a lake is modeled by $P(t) = 1200 \cdot 1.04^t$ , where $t$ is in years. What does the number $1.04$ tell you?
1. a)The population grows by 4% each year
2. b)The population grows by 104% each year
3. c)The population grows by 4 fish each year
4. d)There are 1.04 fish at the start
Answer: The population grows by 4% each year
$1.04 = 1 + 0.04$ , so the population is multiplied by $1.04$ each year: it grows by 4% per year. The starting value is $a = 1200$ fish.
3.
Each list shows the values of a quantity for $t = 0, 1, 2, 3$ . Which list follows an exponential model?
1. a)5; 10; 15; 20
2. b)5; 8; 11; 14
3. c)5; 10; 20; 40
4. d)5; 9; 13; 17
Answer: 5; 10; 20; 40
In $5; 10; 20; 40$ every value is $2$ times the previous one (constant ratio), so it is exponential. The other lists add the same amount each time ( $+3$ , $+5$ , $+4$ ), so they are linear.
4.
You invest 2000 dollars at 6% annual interest, compounded quarterly. Which model gives the amount $A$ after $t$ years?
1. a) $A = 2000 \cdot 1.015^{t}$
2. b) $A = 2000 \cdot 1.06^{4t}$
3. c) $A = 2000 \cdot 1.24^{t}$
4. d) $A = 2000 \cdot 1.015^{4t}$
Answer: $A = 2000 \cdot 1.015^{4t}$
$A = 2000\left(1 + \frac{0.06}{4}\right)^{4t} = 2000 \cdot 1.015^{4t}$ . The first choice applies the yearly rate every quarter, the third forgets that there are $4t$ quarters, and $1.24$ is not a factor of this account.
5.
A car bought for 20000 dollars loses 10% of its value every year. Four students computed its value after 4 years. Which result is correct?
1. a)12000
2. b)29282
3. c)2
4. d)13122
Answer: 13122
$20000 \cdot 0.9^4 = 20000 \cdot 0.6561 = 13122$ . $29282$ used $1.1$ (growth), $12000$ subtracted $2000$ every year (linear), and $2$ used the factor $0.1$ .
6.
A video has 500 views, and the number of views doubles every day. How many views does it have after 6 days?
Answer: 32000
$V(6) = 500 \cdot 2^6 = 500 \cdot 64 = 32000$ views.
7.
A substance has a half-life of 5 hours. A sample starts with 80 mg. How many mg are left after 20 hours?
Answer: 5
$\frac{20}{5} = 4$ half-lives: $80 \cdot \left(\frac{1}{2}\right)^4 = \frac{80}{16} = 5$ mg ( $80 \to 40 \to 20 \to 10 \to 5$ ).
8.
You deposit 1000 dollars at 8% annual interest, compounded quarterly. How much is in the account after 5 years? Round to the nearest cent.
Answer: 1485.95
$A = 1000 \cdot 1.02^{20} \approx 1000 \cdot 1.485947 \approx 1485.95$ dollars.
9.
1500 dollars are invested at 5% per year, compounded continuously. Use $e^{0.5} \approx 1.6487$ . How much is the investment worth after 10 years? Round to the nearest dollar.
Answer: 2473
$A = 1500 \cdot e^{0.5} \approx 1500 \cdot 1.6487 = 2473.05$ , so about 2473 dollars.
10.
A table gives $y = 80$ at $t = 0$ , $y = 120$ at $t = 1$ , $y = 180$ at $t = 2$ and $y = 270$ at $t = 3$ . Decide whether it is linear or exponential, then find the exact value of $y$ at $t = 5$ .
Answer: 607.5
Differences $40, 60, 90$ are not constant, but every ratio is $1.5$ : exponential, $y = 80 \cdot 1.5^t$ . Then $y(5) = 80 \cdot 1.5^5 = 80 \cdot 7.59375 = 607.5$ .
11.
A laptop costs 1200 dollars and loses 20% of its value every year. What is it worth after 3 years?
Answer: 614.4
1. What is the decay factor $b$ per year? 0.8
2. Compute $0.8^3$ . 0.512
3. What is the value after 3 years? 614.4
12.
On Monday, 4 students know a new dance challenge. Every day the number of students who know it triples. How many students know it 4 days later?
Answer: 324
1. What is the growth factor $b$ per day? 3
2. Compute $3^4$ . 81
3. How many students know the challenge after 4 days? 324
13.
You invest 3000 dollars at 5% annual interest, compounded semiannually (twice a year). How much do you have after 6 years? Round to the nearest cent.
Answer: 4034.67
1. What is the rate per half-year, $\frac{r}{n}$ , as a decimal? 0.025
2. How many half-years are there in 6 years ( $nt$ )? 12
3. Compute $3000 \cdot 1.025^{12}$ to the nearest cent. 4034.67
14.
A radioactive material has a half-life of 5 years. A sample has 640 g. How many grams are left after 15 years?
Answer: 80
1. How many half-lives pass in 15 years ( $\frac{t}{T}$ )? 3
2. What is $\left(\frac{1}{2}\right)^3$ as a decimal? 0.125
3. How many grams are left? 80
15.
A patient gets 500 mg of a medicine. The amount in the blood decays continuously at a rate of 10% per hour, so $A(t) = 500 \cdot e^{-0.1t}$ . Use $e^{-0.2} \approx 0.8187$ . How many mg are left after 2 hours? Round to the nearest mg.
Answer: 409
1. What is the exponent $rt$ for $t = 2$ ? -0.2
2. Which value do you use for $e^{-0.2}$ (4 decimal places)? 0.8187
3. How many mg are left after 2 hours (nearest mg)? 409
16.
Bank A pays 5% per year, compounded annually. Bank B pays 5% per year, compounded monthly. You deposit 10000 dollars for 3 years. How much more does Bank B pay? Round every amount to the nearest cent.
Answer: 38.47
1. How much does Bank A pay out after 3 years? 11576.25
2. How much does Bank B pay out after 3 years? 11614.72
3. How much more does Bank B pay? 38.47
17.
A town had 2000 inhabitants. Two years later it had 2880. Assume exponential growth $P(t) = 2000 \cdot b^t$ ( $t$ in years). Find $b$ , the yearly growth rate, and the population after 4 years (rounded to whole inhabitants).
Answer: 4147
1. What is $b^2 = \frac{2880}{2000}$ ? 1.44
2. What is $b$ (positive)? 1.2
3. What is the yearly growth rate in percent (write only the number)? 20
4. Compute $P(4) = 2000 \cdot 1.2^4$ and round to whole inhabitants. 4147
18.
The value of a car is $V(t) = 18000 \cdot 0.88^t$ dollars, $t$ years after it was bought. Interpret the model, then find after how many whole years the value first drops below 10000 dollars.
Answer: 5
1. What was the purchase price of the car? 18000
2. By what percent does the value drop each year (write only the number)? 12
3. What is $V(2) = 18000 \cdot 0.88^2$ ? 13939.2
4. After how many whole years is the value below 10000 for the first time? 5