Exponential Growth and Decay Models

Two numbers make the model

$$y = a \cdot b^x$$

Every exponential model is built by finding $a$ and $b$. Everything else is arithmetic.

Symbol	What to look for
$a$	the amount at the start, before any time passes
$b$	what one time step multiplies by
$x$	how many time steps, in the same unit $b$ was measured in

The last column matters. If $b$ is a yearly multiplier, $x$ counts years. Mixing units is the error that produces answers off by orders of magnitude.

From a description

A percent per time step converts straight to $b$:

$$\text{growth: } b = 1 + r \qquad \text{decay: } b = 1 - r$$

Situation	$a$	$b$	Model
$2,000 growing $6\%$ a year	$2000$	$1.06$	$2000 \cdot 1.06^x$
a town of $8{,}000$ shrinking $2\%$ a year	$8000$	$0.98$	$8000 \cdot 0.98^x$
$40$ mg of medicine, $30\%$ leaves each hour	$40$	$0.70$	$40 \cdot 0.70^x$
bacteria doubling every hour, starting at $500$	$500$	$2$	$500 \cdot 2^x$

“Doubling” is $b = 2$, “tripling” is $b = 3$, and “halving” is $b = 0.5$. These are percent changes stated a different way: doubling is a $100\%$ increase.

From a table

Divide each output by the one before it. A constant answer is $b$.

$x$	$y$
$0$	$600$
$1$	$540$
$2$	$486$
$3$	$437.4$

$$\frac{540}{600} = 0.9 \qquad \frac{486}{540} = 0.9 \qquad \frac{437.4}{486} = 0.9$$

So $b = 0.9$ and $a = 600$, giving $y = 600 \cdot 0.9^x$. The $0.9$ says the amount falls by $10\%$ each step.

If the table does not start at $x = 0$, work backwards to find $a$ by dividing once per step.

From two points

A culture has $300$ cells at hour $2$ and $2{,}400$ at hour $5$.

The outputs are $3$ steps apart, so $b^3$ is the ratio between them:

$$b^3 = \frac{2400}{300} = 8 \quad\Rightarrow\quad b = \sqrt[3]{8} = 2$$

Then work back to $x = 0$ by dividing twice:

$$a = \frac{300}{2^2} = 75$$ $$y = 75 \cdot 2^x$$

Check it against a given point. At $x = 5$: $75 \cdot 32 = 2400$ ✓

Interpreting the parts in context

A model is only useful if its numbers can be read back into the situation.

A savings account is modeled by $A = 1500 \cdot 1.045^t$, with $t$ in years.

Part	Reads as
$1500$	the amount deposited
$1.045$	each year keeps everything and adds $4.5\%$
$0.045$	the annual interest rate
$t$	years since the deposit

The question “how much interest in year 3?” is not the same as “what is the balance after 3 years?”, and reading the model carefully is what separates them.

Compounding more often divides the rate and multiplies the exponent:

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

with $n$ periods a year. Monthly compounding at $6\%$ uses $1 + \tfrac{0.06}{12}$ twelve times a year, which beats $1.06$ once a year — slightly.

Why the curve never reaches zero

• growth
• decay

Feature	What it says
$y$-intercept	the start value $a$
rising or falling	whether $b$ is above or below $1$
steepness	how far $b$ is from $1$
flattening toward the $x$-axis	the horizontal asymptote at $y = 0$

The curve never reaches zero. Multiplying a positive number by a positive factor always leaves something positive, however many times you do it. In a decay model that means the substance thins out forever rather than disappearing at a particular moment, which is why half-life is quoted instead of a lifetime.

Worked examples

Common mistakes

Practice problems