When do I use $e^{rt}$ instead of $(1 + r)^t$?

Use $e^{rt}$ only when the problem says the growth or decay is **continuous** (or compounded continuously). If it says yearly, monthly, quarterly or gives no compounding at all, use $(1 + r)^t$ or $\left(1 + \frac{r}{n}\right)^{nt}$.

My calculator has no $e$ key. What do I do?

Use the value given in the problem, like $e^{0.4} \approx 1.4918$, or type $2.71828$ for $e$. The calculator in this lesson has $e$, so there you can type e^0.4 directly. With 5 decimals of $e$ your answer will round to the same cent or whole number in the problems of this lesson.

Why does the 5th step matter if the computation is correct?

Most wrong answers come from the setup, not the arithmetic: a factor of $0.15$ instead of $0.85$, or $t$ instead of $\frac{t}{2}$. Asking "should this be bigger or smaller than the start?" catches those errors in a few seconds.

Teacher Resources

Exponential Function Word Problems

Solve real exponential word problems with one 5-step method: compound interest, depreciation, viral spread, half-life and continuous growth.

Advanced35 minLesson

Printable worksheet

Definition

An exponential word problem describes a quantity that changes by the same factor in every equal time period. Your job is to turn the story into a model of the form

y = a \cdot b^{t}

where

a

is the initial value and

b

is the growth factor (

b > 1

) or decay factor (

0 < b < 1

The 5-step method works for every problem in this lesson:

1.Read: find the starting amount, the rate or factor, the time and the units.
2.Identify: is it growth or decay? Find the factor $b$ (for a rate $r$ : $b = 1 + r$ or $b = 1 - r$ ).
3.Write the model: choose the right formula and put in the numbers.
4.Compute: evaluate with a calculator and round as the problem asks.
5.Check: is the answer reasonable (bigger or smaller than the start?) and in the right units?

Formula recap (from the earlier lessons):

Compound interest, $n$ times per year: $A = P\left(1 + \frac{r}{n}\right)^{nt}$
Continuous growth or decay: $A = P \cdot e^{rt}$
Half-life $T$ : $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$

Try it now

A phone loses 15% of its value every year. Which factor

b

do you use in the model

V(t) = a \cdot b^t

Worked Examples

You deposit 5000 dollars in a savings account that pays 6% annual interest, compounded monthly. How much is in the account after 10 years?

Read: list the quantities

$P = 5000$ dollars, $r = 0.06$ , $n = 12$ (monthly), $t = 10$ years → Start, rate, periods per year and time found

Identify the factor per month

$1 + \frac{r}{n} = 1 + \frac{0.06}{12} = 1.005$ → Growth, factor $1.005$ per month

Write the model

$A = 5000 \cdot 1.005^{12 \cdot 10} = 5000 \cdot 1.005^{120}$ → $A = 5000 \cdot 1.005^{120}$

Compute

$1.005^{120} \approx 1.819397$ , so $A \approx 5000 \cdot 1.819397 \approx 9096.98$ → $A \approx 9096.98$

Check

Yearly compounding gives $5000 \cdot 1.06^{10} \approx 8954.24$ . Monthly should give a little more, and it does. → Reasonable, in dollars

Answer: After 10 years the account holds about 9096.98 dollars.

Common Mistakes

Using $r$ and $t$ instead of $\frac{r}{n}$ and $nt$ in compound interest

Why it's wrong: With monthly compounding the bank pays $\frac{6\%}{12} = 0.5\%$ each month, 120 times in 10 years. Writing $5000 \cdot 1.06^{120}$ or $5000 \cdot 1.005^{10}$ mixes up the period.

Correct: Rate per period $\frac{r}{n}$ , number of periods $nt$ : $A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \cdot 10}$ .

Using the rate as the factor, for example $b = 0.15$ for a 15% loss

Why it's wrong: A 15% loss means 85% of the value stays. $24000 \cdot 0.15^5$ is almost zero, which no car does in 5 years.

Correct: Decay: $b = 1 - r = 0.85$ . Growth: $b = 1 + r$ . The factor is what is left (or what you have) after one period.

Using $t$ instead of $\frac{t}{T}$ when the period is not 1

Why it's wrong: "Triples every 2 hours" or "half-life of 6 hours" means the factor applies once per 2 or 6 hours, not every hour. $40 \cdot 3^{10}$ is far too big.

Correct: Count the periods first: exponent $= \frac{\text{time}}{\text{length of one period}}$ , for example $3^{10/2} = 3^5$ .

Skipping the reasonableness check

Why it's wrong: A decay answer larger than the start, or a growth answer smaller than the start, always signals a wrong factor or a wrong exponent.

Correct: Step 5 every time: growth ends above $a$ , decay ends between $0$ and $a$ , and the units match the question.

Interactive Visual

Interest Calculator

Principal ($5,000.00)

Interest Rate (6%)

Time Period (10 years)

Compounding

Formula

Compound Interest:A = P(1 + r/n)^(nt)A = 5000 · (1 + 0.06/12)^(12 · 10) = $9,096.98

Compound Interest

Interest Earned:$4,096.98

Total Amount:$9,096.98

Same money, different compounding

Compounding	n	Total Amount	Interest Earned
Annually	1	$8,954.24	$3,954.24
Monthly	12	$9,096.98	$4,096.98
Daily	365	$9,110.14	$4,110.14
Continuously	∞	$9,110.59	$4,110.59

Growth Over Time

Sequence Explorer

First Term (a₁ = 200)

Common Ratio (r = 0.5)

200,

100,

50,

25,

12.5,

6.25

, ...

Formula

a_n = 200 × 0.5^(n-1)

a_1

200

a_5

12.5

a_10

0.391

a_20

0.000381

Interactive Sandbox

Expression Calculator

Try these:

History

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Practice Problems

18 problems

Problem 1 of 18

Easy

A phone loses 15% of its value every year. Which factor $b$ do you use in the model $V(t) = a \cdot b^t$ ?

Practice Problems: Exponential Function Word Problems

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?
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?
3. Each list shows the values of a quantity for t = 0, 1, 2, 3. Which list follows an exponential model?
4. You invest 2000 dollars at 6% annual interest, compounded quarterly. Which model gives the amount A after t years?
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?
6. A video has 500 views, and the number of views doubles every day. How many views does it have after 6 days?
7. A substance has a half-life of 5 hours. A sample starts with 80 mg. How many mg are left after 20 hours?
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.
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.
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.
11. A laptop costs 1200 dollars and loses 20% of its value every year. What is it worth after 3 years?
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?
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.
14. A radioactive material has a half-life of 5 years. A sample has 640 g. How many grams are left after 15 years?
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.
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.
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).
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.

Why It Matters

Exponential models are the language of money, medicine, technology and nature. Knowing how to set them up yourself means you can check claims instead of just believing them:

Savings and loans: the difference between yearly and monthly compounding is real money over 10 or 20 years
Buying a car or a phone: depreciation tells you what it will be worth when you sell it
Health: half-life tells a doctor how much medicine is still in your body after a few hours
Social media: viral posts grow by a factor in each time period, which is why they explode so fast
Science: populations, radioactive materials and cooling objects all follow exponential laws

This lesson brings together everything from the exponential functions unit. The 5-step method is the same every time; only the formula in step 3 changes.

Real World Applications

Saving for University

Banks compound interest yearly, quarterly, monthly or daily. The formula $A = P\left(1 + \frac{r}{n}\right)^{nt}$ lets you compare offers before you choose.

Example:

1000 dollars at 4% compounded quarterly for 3 years: $1000 \cdot 1.01^{12} \approx 1126.83$ dollars.

1Try It Yourself

You put 4000 dollars into an account paying 6% annual interest, compounded quarterly, for 2 years.

How much money is in the account after 2 years (to the nearest cent)?

Step 1: Write the mathematical expression

Use $A = P\left(1 + \frac{r}{n}\right)^{nt}$ with $P = 4000$ , $r = 0.06$ , $n = 4$ , $t = 2$ :

Bacteria in the Lab

Microbiologists model growth that happens all the time, not in steps, with continuous growth $A = P \cdot e^{rt}$.

Example:

A culture of 1000 cells growing continuously at 2% per hour for 10 hours: $1000 \cdot e^{0.2} \approx 1000 \cdot 1.2214 \approx 1221$ cells.

2Try It Yourself

A culture starts with 800 cells and grows continuously at a rate of 3% per hour. Use $e^{0.3} \approx 1.3499$ .

About how many cells are there after 10 hours (to the nearest whole cell)?

Step 1: Write the mathematical expression

Use $A = P \cdot e^{rt}$ with $P = 800$ , $r = 0.03$ , $t = 10$ :

Caffeine in Your Body

Doctors and pharmacists use half-life to decide how often a medicine is taken. Caffeine has a half-life of about 5 hours in an adult.

Example:

A 100 mg coffee at 8:00 still leaves $100 \cdot \left(\frac{1}{2}\right)^{1} = 50$ mg at 13:00.

3Try It Yourself

An energy drink contains 160 mg of caffeine. Assume a half-life of 5 hours.

How much caffeine is left after 10 hours?

Step 1: Write the mathematical expression

Use $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$ with $N_0 = 160$ , $T = 5$ , $t = 10$ :

Key Takeaways

1Every exponential word problem fits $y = a \cdot b^{t}$ : $a$ is the starting value, $b$ the factor per period
2Use the 5 steps: read, identify growth or decay and the factor, write the model, compute, check
3A rate $r$ gives the factor $b = 1 + r$ for growth and $b = 1 - r$ for decay
4Compound interest: $A = P\left(1 + \frac{r}{n}\right)^{nt}$ ; continuous: $A = P \cdot e^{rt}$ ; half-life: $N_0 \left(\frac{1}{2}\right)^{t/T}$
5When the factor applies once every $T$ units of time, the exponent is $\frac{t}{T}$
6In a table, constant differences mean linear, constant ratios mean exponential

Frequently Asked Questions

Look for words like "percent per year", "doubles", "triples", "halves" or "half-life": the change is a factor. If the story says "increases by 50 each year", the change is an amount, so it is linear. In a table, test the ratios of consecutive values.

Glossary

Exponential model: A formula $y = a \cdot b^{t}$ that describes a quantity changing by the same factor $b$ in every equal time period
Growth factor / decay factor: The number $b$ you multiply by each period: $b = 1 + r$ for growth, $b = 1 - r$ for decay
Compound interest: Interest added to the balance $n$ times per year, so later interest is also earned on earlier interest
Depreciation: The loss of value of an object (car, phone, machine) over time, often a fixed percentage per year
Half-life: The time $T$ it takes for a decaying quantity to fall to half of its value
Continuous growth: Growth that happens at every moment, modeled with $A = P \cdot e^{rt}$

Formula Card

General Exponential Model

y = a \cdot b^{t}

$a$ = initial value, $b$ = factor per period ($b = 1 + r$ growth, $b = 1 - r$ decay), $t$ = number of periods.

Compound Interest

A = P\left(1 + \frac{r}{n}\right)^{nt}

$P$ = starting amount, $r$ = annual rate as a decimal, $n$ = compounding periods per year, $t$ = years.

Continuous Growth or Decay

A = P \cdot e^{rt}

$r > 0$ for growth, $r < 0$ for decay, $e \approx 2.71828$.

Half-Life

N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}

$N_0$ = starting amount, $T$ = half-life, $t$ = time passed (same unit as $T$).

Exponential Function Word Problems

Definition

Worked Examples

Common Mistakes

Interactive Visual

Interest Calculator

Compound Interest

Same money, different compounding

Growth Over Time

Sequence Explorer

Formula

Interactive Sandbox

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Saving for University

Bacteria in the Lab

Caffeine in Your Body

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

General Exponential Model

Compound Interest

Continuous Growth or Decay

Half-Life

More in This Topic

Introduction to Exponential Functions

Exponential Growth

Graphing Exponential Functions

Exponential Decay