Exponential Functions
Everything from today's session: writing equations from two points, compound interest, exponential decay, and matching graphs. Same problems, same wording, more reps.
The One Formula
Every problem on this page uses the same equation: $y = A \cdot B^x$. The only thing that changes is what they give you and what they're asking for. Sometimes you're finding $A$ and $B$ from two points. Sometimes you already know them and you're just plugging in. Once you see that it's always the same formula, these stop being five different problem types and start being one.
Write the Exponential Function Through Two Points
Two points, two unknowns ($A$ and $B$). The move: divide the two equations to kill $A$, find $B$ first, then plug back in for $A$.
$81 = A \cdot B^3$ and $729 = A \cdot B^5$
$\dfrac{A \cdot B^5}{A \cdot B^3} = \dfrac{729}{81}$
The $A$'s cancel. Subtract exponents: $B^2 = 9$.
$B = 3$.
$81 = A \cdot 3^3 = A \cdot 27$, so $A = 3$.
You can double-check: $3 \cdot 3^5 = 3 \cdot 243 = 729$. ✓
Shortcut — when one point has $x = 0$. If they give you a point like $(0, 5)$, then $B^0 = 1$, so $y = A \cdot 1 = A$. You get $A$ for free — just read it off the $y$-value. Then plug in the other point to find $B$. When neither point has $x = 0$, you have to divide first, like above.
Practice 1. Write the exponential function $y = A \cdot B^x$ that passes through $(0, 5)$ and $(3, 135)$.
Hint ›
One of the points has $x = 0$. What does that give you immediately?
Check it ›
$A = 5$ (from the zero point). Then $135 = 5 \cdot B^3$, so $B^3 = 27$, and $B = 3$.
$y = 5 \cdot 3^x$.
Practice 2. Write the exponential function $y = A \cdot B^x$ that passes through $(2, 48)$ and $(5, 384)$.
Hint ›
Neither point has $x = 0$, so divide. Put the bigger power on top: $\dfrac{384}{48}$. The exponent difference is $5 - 2 = 3$.
Check it ›
$B^3 = \dfrac{384}{48} = 8$, so $B = 2$.
Plug back in: $48 = A \cdot 2^2 = 4A$, so $A = 12$.
$y = 12 \cdot 2^x$.
Practice 3. Write the exponential function $y = A \cdot B^x$ that passes through $(1, 6)$ and $(4, 162)$.
Hint ›
Divide: $\dfrac{162}{6} = 27$. Exponent difference is $4 - 1 = 3$. So $B^3 = 27$.
Check it ›
$B = 3$. Then $6 = A \cdot 3^1 = 3A$, so $A = 2$.
$y = 2 \cdot 3^x$.
Practice 4. Write the exponential function $y = A \cdot B^x$ that passes through $(4, 32)$ and $(9, 1024)$.
Hint 1 ›
Exponent difference: $9 - 4 = 5$. Put the bigger power on top.
Hint 2 ›
$B^5 = \dfrac{1024}{32} = 32$. These are almost always 2 or 3 — try $2^5$.
Check it ›
$B = 2$. Then $32 = A \cdot 2^4 = 16A$, so $A = 2$.
$y = 2 \cdot 2^x$.
Practice 5. Write the exponential function $y = A \cdot B^x$ that passes through $(1, 10)$ and $(3, 250)$.
Hint ›
$B^2 = \dfrac{250}{10} = 25$.
Check it ›
$B = 5$. Then $10 = A \cdot 5$, so $A = 2$.
$y = 2 \cdot 5^x$.
Pattern to notice. The base $B$ in these problems is almost always $2$, $3$, or $5$. If you're stuck on the $n$th root, just try those three. If the number ends in $0$ or $5$, try $5$ first.
Compound Interest
One deposit, sits there for years, earns interest. The formula: $A = P\left(1 + \dfrac{r}{n}\right)^{nt}$
What the letters mean:
$P$ = principal (how much you put in)
$r$ = rate as a decimal (6% → 0.06)
$n$ = number of times it compounds per year (annually = 1)
$t$ = time in years
If they don't say how it compounds, use $n = 1$.
Watch out. Today you accidentally multiplied by $150$ on a problem that started with $640$. The principal changes every problem — read it fresh every time.
Practice 1. $200 is invested at 5% compounded annually for 10 years. Find the balance.
Hint ›
$200 \cdot (1.05)^{10}$. Do the exponent first, then multiply.
Check it ›
$(1.05)^{10} \approx 1.6289$. Multiply: $200 \times 1.6289 = \$325.78$.
Practice 2. $500 is invested at 8% each year for 12 years. Find the balance.
Hint ›
They didn't say "compounded" — use $n = 1$ anyway. $500 \cdot (1.08)^{12}$.
Check it ›
$(1.08)^{12} \approx 2.5182$. Multiply: $500 \times 2.5182 = \$1{,}259.08$.
Practice 3. $300 is invested at 4% compounded annually for 20 years. Find the balance.
Check it ›
$300 \cdot (1.04)^{20} = 300 \times 2.1911 = \$657.34$.
Practice 4. $400 is invested at 3% compounded annually for 25 years. Find the balance.
Check it ›
$400 \cdot (1.03)^{25} = 400 \times 2.0938 = \$837.51$.
Exponential Decay
Same $y = A \cdot B^x$ formula, same divide-to-find-$B$ method from Section 1 — just with a story wrapped around it. "How much is left after this many hours?"
At 0 hours: 1800 mg → $(0, 1800)$
At 80 hours: 450 mg → $(80, 450)$
$450 = 1800 \cdot B^{80}$
$B^{80} = \dfrac{450}{1800} = \dfrac{1}{4}$
$B = \left(\dfrac{1}{4}\right)^{1/80}$ — leave this as a decimal in your calculator.
$y = 1800 \cdot \left(\dfrac{1}{4}\right)^{398/80}$
Calculator: $(0.25)$ raised to the power of $(398 \div 80)$, then multiply by $1800$.
Why 398/80? We know that 80 jumps gives us a decay of $\frac{1}{4}$. To find what happens after 398 jumps, we raise $\frac{1}{4}$ to the power of $\frac{398}{80}$ — that's how many "80-hour chunks" fit into 398 hours. Same idea as the two-point divide: you're scaling the decay.
Practice 1. A scientist begins with 1600 milligrams of a substance. After 60 hours, 400 milligrams remain. How many milligrams will remain after 180 hours?
Hint 1 ›
$A = 1600$. Divide: $\dfrac{400}{1600} = \dfrac{1}{4}$. That happened over 60 hours.
Hint 2 ›
$y = 1600 \cdot \left(\dfrac{1}{4}\right)^{180/60} = 1600 \cdot \left(\dfrac{1}{4}\right)^{3}$
Check it ›
$\left(\frac{1}{4}\right)^3 = \frac{1}{64}$. Then $\frac{1600}{64} = 25$ mg.
Practice 2. A scientist begins with 2700 milligrams of a substance. After 40 hours, 900 milligrams remain. How many milligrams will remain after 120 hours?
Hint ›
$\dfrac{900}{2700} = \dfrac{1}{3}$. How many 40-hour chunks fit in 120?
Check it ›
$y = 2700 \cdot \left(\frac{1}{3}\right)^{120/40} = 2700 \cdot \left(\frac{1}{3}\right)^3 = 2700 \cdot \frac{1}{27} = 100$ mg.
Practice 3. A scientist begins with 3200 milligrams of a substance. After 50 hours, 800 milligrams remain. How many milligrams will remain after 175 hours?
Hint ›
$\dfrac{800}{3200} = \dfrac{1}{4}$. The power is $\frac{175}{50} = 3.5$.
Check it ›
$y = 3200 \cdot \left(\frac{1}{4}\right)^{3.5} = 3200 \cdot \frac{1}{128} = 25$ mg.
Population Growth & Decline
They give you the starting number, the annual rate, and ask what the population will be in $x$ years. No dividing needed — just plug straight into $y = A \cdot B^x$.
Think of it like a sale: 16% off means you pay 84%.
Round down — two-thirds of a person doesn't count.
Growth works the same way. If the population is growing by $10\%$/year, then $B = 1.10$ — you keep everything plus $10\%$ more. Decay: subtract the percent from $1$. Growth: add it to $1$.
Practice 1. The population of a town today is 2,500. For the last several years, the population has declined by 20% every year. If this continues, what will the population be 10 years from now?
Hint ›
$B = 1 - 0.20 = 0.80$. Then $y = 2500 \cdot (0.80)^{10}$.
Check it ›
$(0.80)^{10} \approx 0.10737$. Multiply: $2500 \times 0.10737 \approx 268.4$.
Round down: 268 people.
Practice 2. The population of a town today is 1,500. For the last few years, the population has declined by 12% every year. If this continues, what will the population be 15 years from now?
Check it ›
$B = 0.88$. $y = 1500 \cdot (0.88)^{15} \approx 1500 \times 0.14697 \approx 220.5$.
Round down: 220 people.
Matching Exponential Graphs
When they show you four graphs and ask "which one matches this function?" — don't try to graph the whole thing. Use one of two moves.
Move 1 — plug in $x = 0$. Find the $y$-intercept and match it. Every basic exponential $B^x$ passes through $(0, 1)$ because anything to the zero power is $1$. If there's a coefficient or a shift, that changes.
Move 2 — find the asymptote. A basic exponential approaches $y = 0$ as $x$ goes left. If you subtract $1$ from the function, the asymptote drops to $y = -1$. If you subtract $3$, it drops to $y = -3$. Match the dotted line.
Horizontal shifts. A "$+$" inside the exponent shifts left. $3^{x+3}$ shifts the normal $3^x$ graph $3$ units left. Instead of hitting $y = 1$ at $x = 0$, it hits $y = 1$ at $x = -3$.
Practice 1. What is the $y$-intercept of $y = \left(\frac{1}{3}\right)^{x+1}$?
Check it ›
Plug in $x = 0$: $y = \left(\frac{1}{3}\right)^{0+1} = \frac{1}{3}$. The graph crosses the $y$-axis at $\frac{1}{3}$.
Practice 2. $g(x) = 4^x - 3$ is a transformation of $f(x) = 4^x$. What is the horizontal asymptote of $g(x)$?
Check it ›
The basic $4^x$ approaches $y = 0$. Subtracting $3$ shifts everything down $3$. The asymptote is $y = -3$.
Practice 3. $y = 5^{x+2}$ is a horizontal shift of $y = 5^x$. Where does the shifted graph cross $y = 1$?
Check it ›
The basic graph crosses $y = 1$ at $x = 0$. The "$+2$" inside the exponent shifts left by $2$. It crosses $y = 1$ at $x = -2$.
Practice 4. What is the $y$-intercept of $y = 3 \cdot 2^{x-1}$?
Hint ›
Plug in $x = 0$: $y = 3 \cdot 2^{-1}$. What is $2^{-1}$?
Check it ›
$2^{-1} = \frac{1}{2}$, so $y = 3 \cdot \frac{1}{2} = \frac{3}{2} = 1.5$. The graph crosses the $y$-axis at $1.5$.
All Types — Scrambled
Same problems as above, shuffled. Spot the type first, then solve.
1 · $350 is invested at 6% compounded annually for 18 years. Find the balance.
Check ›
$350 \cdot (1.06)^{18} = 350 \times 2.8543 = \$999.02$.
2 · Write the exponential function through $(2, 36)$ and $(5, 972)$.
Check ›
$B^3 = \frac{972}{36} = 27$, so $B = 3$. Then $36 = A \cdot 9$, so $A = 4$. $y = 4 \cdot 3^x$.
3 · What is the horizontal asymptote of $y = 5^x + 2$?
Check ›
Adding $2$ shifts the graph up by $2$. The asymptote moves from $y = 0$ to $y = 2$.
4 · A town of 3,200 people is declining by 25% every year. What will the population be in 8 years?
Check ›
$B = 0.75$. $y = 3200 \cdot (0.75)^{8} = 3200 \times 0.10011 \approx 320.4$. Round down: 320 people.
5 · A scientist begins with 2400 milligrams of a substance. After 46 hours, 1200 milligrams remain. How many milligrams will remain after 219 hours?
Check ›
$\frac{1200}{2400} = \frac{1}{2}$ over 46 hours. $y = 2400 \cdot \left(\frac{1}{2}\right)^{219/46} \approx 88.5$ mg. Round to nearest: 88 or 89 mg.
6 · Write the exponential function through $(0, 7)$ and $(2, 63)$.
Check ›
$A = 7$ (zero-point shortcut). $63 = 7 \cdot B^2$, so $B^2 = 9$, $B = 3$. $y = 7 \cdot 3^x$.
7 · What is the $y$-intercept of $y = \left(\frac{1}{2}\right)^{x+3}$?
Check ›
Plug in $x = 0$: $y = \left(\frac{1}{2}\right)^{3} = \frac{1}{8}$.
8 · In your own words: when a problem says a population "declines by 16% every year," why do you use $B = 0.84$ and not $B = 0.16$?
Check ›
$B$ is the multiplier — what fraction of the population stays each year, not what leaves. If $16\%$ leaves, $84\%$ stays. Think: 16% off means you pay 84%.
9 · Write the exponential function through $(3, 24)$ and $(6, 192)$.
Check ›
$B^3 = \frac{192}{24} = 8$, so $B = 2$. Then $24 = A \cdot 2^3 = 8A$, so $A = 3$. $y = 3 \cdot 2^x$.
10 · $750 is invested at 9% each year for 8 years. Find the balance.
Check ›
$750 \cdot (1.09)^{8} = 750 \times 1.9926 = \$1{,}494.43$.
Before You Leave — Cheat-Sheet Lines
"$y = A \cdot B^x$ — two points: divide to find $B$ (bigger power on top), then plug back in for $A$. If one point has $x = 0$, read $A$ straight off.
Compound interest: $A = P(1 + r/n)^{nt}$ — exponent first, then multiply by $P$.
Decay/growth rate to $B$: subtract the decay% from 1 (or add the growth%). 16% decline → $B = 0.84$."