Model compound interest, population growth, half-life, and depreciation using exponential functions. Write equations from word descriptions, evaluate them, and interpret outputs in context.
A lesson of Algebra 1, in Exponents & Exponential Functions, about 35 minutes.
Exponential Applications
By the end of this lesson, you will be able to build an exponential model from a description, handle compound interest and half-life correctly, get the exponent's units right, and say what each part of your model means.
The Same Four Questions Every Time
Whatever the context, building the model is the same procedure.
What is the starting amount? That is $a$
Is it growing or shrinking? That decides which side of $1$ the factor sits
By what factor per period? That is $b$
How many periods? That is the exponent, and it is rarely the raw number in
the problem
$$f(t) = a \cdot b^t$$
Step 4 is where the marks are lost. A half-life of $8$ days over $32$ days is
$t = 4$, not $32$. Interest compounded monthly for $3$ years is $t = 36$, not $3$.
The exponent counts PERIODS, and the period is defined by whatever the rate
refers to.
Always say what $t$ means before computing anything. "Let $t$ be the number of
months" takes four seconds and prevents the only structural error in the topic.
Sort It: What Is the Exponent
Read what the rate refers to, then count periods.
Compound Interest
$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$
$P$ is the principal, what you started with
$r$ is the annual rate as a decimal
$n$ is how many times a year it compounds
$t$ is the number of YEARS
The formula looks unlike $a \cdot b^x$ and is not. $\frac{r}{n}$ is the rate per
period and $nt$ is the number of periods, so it is $a \cdot b^x$ with both parts
spelled out for you.
$$$1000 \text{ at } 6% \text{ compounded quarterly for } 5 \text{ years}$$
The gains shrink each time and never pass a ceiling of about $$1349.86$. That
limit is $Pe^{rt}$, continuous compounding, and it is one of the places $e$ comes
from.
Example 1: Compound Interest
Here is every step, worked all the way through.
$$2000$ at $8%$ compounded annually for $3$ years.
What is P? 2000
Compounded annually, so what is the factor b? 1.08
What is the exponent? 3
What is the balance after 1 year, in dollars? 2160
What is the balance after 3 years, to the nearest dollar? 2519
Half-Life
$$A = A_0\left(\frac{1}{2}\right)^{t/h}$$
where $h$ is the half-life and $t$ is elapsed time in the same units.
$\frac{t}{h}$ is the number of halvings, which is why the formula looks more
complicated than it is: it is simply converting elapsed time into periods.
It never reaches zero, which is not a technicality. Carbon dating works
precisely because a measurable trace remains after thousands of years: with a
half-life of $5730$ years, a sample from $17{,}190$ years ago still holds an
eighth of its original carbon-14, and that eighth is what gets measured.
The same shape describes medication clearing the bloodstream, which is why dosing
schedules are built around a drug's half-life rather than around when it "wears
off".
Worked Example 1: Half-Life
A $240$ mg dose with a $4$-hour half-life, after $12$ hours.
Population and Depreciation
The two everyday cases, and they are the same formula pointed in opposite
directions.
$$P(t) = P_0(1 + r)^t \qquad \textbf{growth}$$
$$V(t) = V_0(1 - r)^t \qquad \textbf{decay}$$
A town of $12{,}000$ growing $2.5%$ a year:
$P(t) = 12000(1.025)^t$, so after $10$ years, about $15{,}361$.
A $$30{,}000$ car losing $18%$ a year:
$V(t) = 30000(0.82)^t$, so after $5$ years, about $$11{,}120$.
Depreciation is not the same as losing a fixed amount. Losing $18%$ of
$$30{,}000$ is $$5{,}400$ in year one and $$4{,}428$ in year two, because the
percentage applies to what remains. A car losing $$5{,}400$ every year would be
worthless in under six; the real one still has value after twenty.
A rate and its reverse do not cancel. Grow $$100$ by $20%$ and then shrink
it by $20%$ and you have $$96$, not $$100$, because the second percentage
applies to the larger amount. That surprises people and it is worth checking once
with numbers.
Worked Example 2: Depreciation
A $$18{,}000$ car loses $20%$ of its value each year.
Worked Example 3: Writing the Model
Bacteria triple every $6$ hours, starting from $400$. How many after $24$ hours?
Interpreting What You Built
A model is only useful if you can say what its parts mean, and problems ask for
this directly.
$$V(t) = 25000(0.85)^t$$
$25000$ is the value at $t = 0$, in dollars
$0.85$ means $85%$ of the value is retained each year
$1 - 0.85 = 0.15$ means it loses $15%$ a year
$t$ is measured in years
Answer in a sentence with units. "About $$11{,}120$ after five years" is an
answer; "$11120$" is a number, and half the applied questions in this topic ask
for the interpretation rather than the arithmetic.
Round money to cents, populations to whole individuals, and only at the very end.
Rounding $1.025$ to $1.03$ at the start compounds into a large error over twenty
periods.
Common Mistakes
Mistake 1: The wrong exponent units
Half-life $4$ hours over $12$ hours is $t = 3$, not $12$.
Mistake 2: Using the rate as the factor
$8%$ growth is $b = 1.08$, not $0.08$.
Mistake 3: Treating depreciation as a fixed loss
The percentage applies to what remains, so less is lost each year.
Mistake 4: Forgetting $n$ in compound interest
Quarterly for $5$ years is $20$ periods at $\frac{r}{4}$ each.
Mistake 5: Rounding early
Round once, at the end.
Where This Goes Next
Every question in this lesson asked "how much after a given time". The reverse
question is far more useful and cannot be answered with the algebra you have.
The unknown is stuck in the exponent, and no amount of dividing or rooting will
bring it down. Logarithms are the inverse operation built for this:
$$t = \log_{1.05}(2) \approx 14.2 \text{ years}$$
Once you have them, "when will the population reach a million", "how old is this
fossil" and "how long until the drug clears" all become one-line calculations.
Bankers approximate it without logs at all. The rule of 72 says a quantity
doubles in about $\frac{72}{r}$ periods, where $r$ is the percentage rate: at
$6%$, roughly $12$ years. It is accurate to within a few percent across the whole
range people actually use, and it comes from $\ln(2) \approx 0.693$ with a
convenient rounding to a number divisible by many things.
Worth carrying: at $7%$ your money doubles every decade, and any quantity growing
at a steady percentage has a fixed doubling time. That single fact makes long-run
exponential behavior intuitive in a way that the formula alone does not.
Check Your Understanding
Five of them, no instructions this time.
Summary
Four questions: starting amount, growing or shrinking, factor per period, and
number of PERIODS.
The exponent counts periods, not raw time. That is where the marks go.
$$A = P\left(1 + \frac{r}{n}\right)^{nt} \qquad A = A_0\left(\frac{1}{2}\right)^{t/h}$$
Both are $a \cdot b^x$ with the period arithmetic written out.
Percentage decay is not a fixed loss, so less is lost each period, and a rise
followed by an equal fall does not return to the start.
Interpret the model in a sentence with units, and round only at the end.