Exponential Functions: Growth and Decay — SAT Math

Introduction

Exponential functions appear in roughly 3–5 questions per SAT Math section, sitting inside the Advanced Math domain (≈35% of the test). This note covers the full scope of the exponential testing point: building the model $f(x) = a \cdot b^x$ from context, telling growth from decay, using doubling and half-life templates, and rewriting an exponential to expose a rate for a different time unit. Interpreting what those models mean in context is covered in the sibling note Interpreting Nonlinear Models in Context.

Core Concept

Every basic exponential function has the form

f(x) = a \cdot b^x

where:

$a$ is the initial value (the output when $x = 0$ , since $b^0 = 1$ ).
$b$ is the growth/decay factor — a constant multiplier applied once per period.

Growth vs. Decay

Condition	Name	Each period…
$b > 1$	Exponential growth	output increases
$0 < b < 1$	Exponential decay	output decreases
$b \le 0$	Not a valid exponential model	—

From a Percent Rate to a Factor

If a quantity changes by $r\%$ per period, convert to a decimal rate $r_d = r/100$ :

b = 1 + r_d \quad (\text{growth}), \qquad b = 1 - r_d \quad (\text{decay})

A 12% annual increase → $b = 1.12$ . A 5% annual decrease → $b = 0.95$ .

Doubling and Half-Life Models

Context	Model
Doubles every $d$ periods	$f(t) = a \cdot 2^{t/d}$
Half-life of $h$ periods	$f(t) = a \cdot \left(\tfrac{1}{2}\right)^{t/h}$

Both are just $a \cdot b^x$ in disguise: writing $2^{t/d} = \left(2^{1/d}\right)^t$ matches the standard form with $b = 2^{1/d}$ .

Rewriting for a Different Time Unit

This is a high-value SAT skill. If $f(t)$ uses one time unit and you need to express it in another, substitute and simplify using exponent rules.

Key idea

Key idea: if $t_1 = k \cdot t_2$ (e.g., 1 year = 12 months, so $t_{\text{months}} = 12 \cdot t_{\text{years}}$ ), then

a \cdot b^{t_{\text{months}}} = a \cdot b^{12 \cdot t_{\text{years}}} = a \cdot \left(b^{12}\right)^{t_{\text{years}}}

The new factor $b^{12}$ is the annual growth factor. Going the other direction (annual → monthly) raises $b$ to the $1/12$ power.

Quick illustration: $f(m) = 500 \cdot (1.03)^m$ with $m$ = months.
To express in years $y$ (where $m = 12y$ ):

f = 500 \cdot (1.03)^{12y} = 500 \cdot \underbrace{(1.03)^{12}}_{\approx\,1.426}\vphantom{)}^y

The annual growth factor is $(1.03)^{12}$ , revealing roughly a 42.6% annual increase.

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