Compound Interest, Monthly

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**.

Depreciation of a Car

A new car costs 24000 dollars and loses 15% of its value every year. What is it worth after 5 years? Round to the nearest cent.

Read: list the quantities: $a = 24000$ dollars, rate $15\% = 0.15$ per year, $t = 5$ years = Start, rate and time found

Identify: growth or decay?: The value goes down, so it is decay: $b = 1 - 0.15 = 0.85$ = Decay factor $0.85$

Write the model: $V(t) = 24000 \cdot 0.85^{t}$, so $V(5) = 24000 \cdot 0.85^{5}$ = $V(5) = 24000 \cdot 0.85^5$

Compute: $0.85^5 \approx 0.443705$, so $V(5) \approx 10648.93$ = $V(5) \approx 10648.93$

Check: The value is below 24000 and above 0. A linear loss of $3600$ per year would give only $6000$: exponential decay loses less each year because 15% of a smaller value is smaller. = Reasonable, in dollars

Answer: After 5 years the car is worth about **10648.93 dollars**.

A Post Goes Viral

At 12:00, a funny video has been shared 40 times. The number of shares triples every 2 hours. How many shares will there be at 22:00 the same day?

Read: list the quantities: $a = 40$ shares, factor $3$ every $2$ hours, time $t = 10$ hours (12:00 to 22:00) = Start, factor and time found

Identify the factor and the period: Growth with $b = 3$, but per 2 hours, so the exponent is $\frac{t}{2}$ = Growth, $b = 3$ per 2 hours

Write the model: $S(t) = 40 \cdot 3^{t/2}$, so $S(10) = 40 \cdot 3^{10/2} = 40 \cdot 3^5$ = $S(10) = 40 \cdot 3^5$

Compute: $3^5 = 243$, so $S(10) = 40 \cdot 243 = 9720$ = $9720$ shares

Check: 5 tripling periods: $40 \to 120 \to 360 \to 1080 \to 3240 \to 9720$. A whole number of shares, growing fast: makes sense. = Reasonable, a count of shares

Answer: At 22:00 the video has about **9720 shares**.

Medicine and Half-Life

A patient takes 200 mg of a medicine. Its half-life in the body is 6 hours. How much is left after 18 hours?

Read: list the quantities: $N_0 = 200$ mg, half-life $T = 6$ hours, $t = 18$ hours = Start, half-life and time found

Identify: growth or decay?: The amount halves, so it is decay with factor $\frac{1}{2}$ per 6 hours = Decay, factor $\frac{1}{2}$

Write the model: $N(t) = 200 \left(\frac{1}{2}\right)^{t/6}$, so $N(18) = 200 \left(\frac{1}{2}\right)^{3}$ = $N(18) = 200 \cdot \left(\frac{1}{2}\right)^3$

Compute: $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$, so $N(18) = \frac{200}{8} = 25$ = $25$ mg

Check: Halve three times: $200 \to 100 \to 50 \to 25$. Less than the dose, never negative, in mg. = Reasonable, in mg

Answer: After 18 hours, **25 mg** of the medicine is left.

Linear or Exponential? Reading a Table

A channel had these subscriber counts: week 0: 500, week 1: 600, week 2: 720, week 3: 864. Is the pattern linear or exponential? Predict the count in week 5.

Check the differences: $600 - 500 = 100$, $720 - 600 = 120$, $864 - 720 = 144$ = Differences are not constant: not linear

Check the ratios: $\frac{600}{500} = 1.2$, $\frac{720}{600} = 1.2$, $\frac{864}{720} = 1.2$ = Constant ratio $b = 1.2$: exponential

Write the model: $a = 500$ (week 0), $b = 1.2$: $S(t) = 500 \cdot 1.2^{t}$ = $S(t) = 500 \cdot 1.2^t$

Compute week 5: $S(5) = 500 \cdot 1.2^5 = 500 \cdot 2.48832 = 1244.16$ = About $1244$ subscribers

Interpret and check: $a = 500$ is the starting count, $b = 1.2$ means 20% growth per week. A linear guess ($+100$ per week) would give only $1000$. = Reasonable, a whole number of people

Answer: The pattern is **exponential** ($S(t) = 500 \cdot 1.2^t$, 20% per week), and in week 5 the channel has about **1244 subscribers**.

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

Why: 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 $nt$: $A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \cdot 10}$.

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

Why: 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.

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

Why: "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 $3^{10/2} = 3^5$.

Mistake: Skipping the reasonableness check

Why: 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.

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.

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

Bacteria in the Lab

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

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.

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.

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

Q: How do I know whether a problem is exponential and not linear?

A: 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.

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

A: 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 $\left(1 + \frac{r}{n}\right)^{nt}$.

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

A: 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.

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

A: 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.