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 min18 problems
Practice Problems

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⋅bty = a \cdot b^{t}
where aa is the initial value and bb is the growth factor (b>1b > 1) or decay factor (0<b<10 < b < 1).
The 5-step method works for every problem in this lesson:
  1. 1.Read: find the starting amount, the rate or factor, the time and the units.
  2. 2.Identify: is it growth or decay? Find the factor bb (for a rate rr: b=1+rb = 1 + r or b=1−rb = 1 - r).
  3. 3.Write the model: choose the right formula and put in the numbers.
  4. 4.Compute: evaluate with a calculator and round as the problem asks.
  5. 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, nn times per year: A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}
  • Continuous growth or decay: A=P⋅ertA = P \cdot e^{rt}
  • Half-life TT: N(t)=N0(12)t/TN(t) = N_0 \left(\frac{1}{2}\right)^{t/T}

Try it now

A phone loses 15% of its value every year. Which factor bb do you use in the model V(t)=a⋅btV(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?
1

Read: list the quantities

P=5000P = 5000 dollars, r=0.06r = 0.06, n=12n = 12 (monthly), t=10t = 10 years → Start, rate, periods per year and time found
2

Identify the factor per month

1+rn=1+0.0612=1.0051 + \frac{r}{n} = 1 + \frac{0.06}{12} = 1.005 → Growth, factor 1.0051.005 per month
3

Write the model

A=5000⋅1.00512⋅10=5000⋅1.005120A = 5000 \cdot 1.005^{12 \cdot 10} = 5000 \cdot 1.005^{120} → A=5000⋅1.005120A = 5000 \cdot 1.005^{120}
4

Compute

1.005120≈1.8193971.005^{120} \approx 1.819397, so A≈5000⋅1.819397≈9096.98A \approx 5000 \cdot 1.819397 \approx 9096.98 → A≈9096.98A \approx 9096.98
5

Check

Yearly compounding gives 5000⋅1.0610≈8954.245000 \cdot 1.06^{10} \approx 8954.24. Monthly should give a little more, and it does. → Reasonable, in dollars

Common Mistakes

Using rr and tt instead of rn\frac{r}{n} and ntnt in compound interest

Why it's wrong: With monthly compounding the bank pays 6%12=0.5%\frac{6\%}{12} = 0.5\% each month, 120 times in 10 years. Writing 5000⋅1.061205000 \cdot 1.06^{120} or 5000⋅1.005105000 \cdot 1.005^{10} mixes up the period.

Correct: Rate per period rn\frac{r}{n}, number of periods ntnt: A=5000(1+0.0612)12⋅10A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \cdot 10}.

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

Why it's wrong: A 15% loss means 85% of the value stays. 24000⋅0.15524000 \cdot 0.15^5 is almost zero, which no car does in 5 years.

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

Using tt instead of tT\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⋅31040 \cdot 3^{10} is far too big.

Correct: Count the periods first: exponent =timelength of one period= \frac{\text{time}}{\text{length of one period}}, for example 310/2=353^{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 aa, decay ends between 00 and aa, and the units match the question.

Interactive Visual

Interest Calculator

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

CompoundingnTotal AmountInterest Earned
Annually1$8,954.24$3,954.24
Monthly12$9,096.98$4,096.98
Daily365$9,110.14$4,110.14
Continuously∞$9,110.59$4,110.59

Growth Over Time

yearsAmount

Sequence Explorer

200,
100,
50,
25,
12.5,
6.25
, ...

Formula

a_n = 200 × 0.5^(n-1)
123456
a_1

200

a_5

12.5

a_10

0.391

a_20

0.000381

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Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
A phone loses 15% of its value every year. Which factor bb do you use in the model V(t)=a⋅btV(t) = a \cdot b^t?

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⋅1.0112≈1126.831000 \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(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt} with P=4000P = 4000, r=0.06r = 0.06, n=4n = 4, t=2t = 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⋅e0.2≈1000⋅1.2214≈12211000 \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 e0.3≈1.3499e^{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⋅ertA = P \cdot e^{rt} with P=800P = 800, r=0.03r = 0.03, t=10t = 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⋅(12)1=50100 \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)=N0(12)t/TN(t) = N_0 \left(\frac{1}{2}\right)^{t/T} with N0=160N_0 = 160, T=5T = 5, t=10t = 10:

Key Takeaways

  • 1Every exponential word problem fits y=a⋅bty = a \cdot b^{t}: aa is the starting value, bb the factor per period
  • 2Use the 5 steps: read, identify growth or decay and the factor, write the model, compute, check
  • 3A rate rr gives the factor b=1+rb = 1 + r for growth and b=1−rb = 1 - r for decay
  • 4Compound interest: A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}; continuous: A=P⋅ertA = P \cdot e^{rt}; half-life: N0(12)t/TN_0 \left(\frac{1}{2}\right)^{t/T}
  • 5When the factor applies once every TT units of time, the exponent is tT\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.
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.
Use erte^{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(1 + r)^t or (1+rn)nt\left(1 + \frac{r}{n}\right)^{nt}.
Use the value given in the problem, like e0.4≈1.4918e^{0.4} \approx 1.4918, or type 2.718282.71828 for ee. The calculator in this lesson has ee, so there you can type e^0.4 directly. With 5 decimals of ee your answer will round to the same cent or whole number in the problems of this lesson.
Most wrong answers come from the setup, not the arithmetic: a factor of 0.150.15 instead of 0.850.85, or tt instead of t2\frac{t}{2}. Asking "should this be bigger or smaller than the start?" catches those errors in a few seconds.

Glossary

Exponential model
A formula y=a⋅bty = a \cdot b^{t} that describes a quantity changing by the same factor bb in every equal time period
Growth factor / decay factor
The number bb you multiply by each period: b=1+rb = 1 + r for growth, b=1−rb = 1 - r for decay
Compound interest
Interest added to the balance nn 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 TT 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⋅ertA = P \cdot e^{rt}

Formula Card

General Exponential Model

y=a⋅bty = 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(1+rn)ntA = 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⋅ertA = P \cdot e^{rt}

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

Half-Life

N(t)=N0(12)t/TN(t) = N_0 \left(\frac{1}{2}\right)^{t/T}

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

Today's challengeOne problem, three tries

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