Interpret exponential functions for growth and decay, distinguish exponential from linear change, work with compound growth, and read basic logarithms as inverse exponents (Functions).

What this topic is asking

An exponential function $f(x) = a \cdot b^{x}$ models a quantity that multiplies by a constant factor each step, the engine of growth and decay. The ACT tests interpreting the initial value and growth factor, telling exponential from linear change, handling compound growth, and reading basic logarithms as inverse exponents.

Reading an exponential function

The two parameters carry all the meaning.

Exponential versus linear change

The headline contrast on the ACT is linear versus exponential.

A constant difference means linear; a constant ratio means exponential.

Compound growth and decay

A common application is compound interest and population change: $A = P(1 + r)^{t}$, where $P$ is the initial amount, $r$ the rate per period (as a decimal), and $t$ the number of periods. For \1000$at 5$A = 1000(1.05)^{3} \approx 1157.63$. **Decay** uses$1 - r$: a car worth$\$20{,}000$ losing 15% a year is $V(t) = 20000(0.85)^{t}$. The growth or decay factor is the base; the exponent is the number of periods.

Logarithms as inverse exponents

A logarithm undoes an exponent. The statement $\log_b x = y$ means exactly $b^{y} = x$: "the power that turns $b$ into $x$". So $\log_2 16 = 4$ because $2^{4} = 16$, and $\log_{10} 1000 = 3$ because $10^{3} = 1000$. On the ACT, log questions are usually this direct translation between log and exponential form, plus occasionally the rules $\log(xy) = \log x + \log y$ and $\log(x^{n}) = n\log x$. Reading a log as "what exponent?" handles the great majority of these items.

The shape of an exponential graph

An exponential graph has a characteristic shape worth recognising. For growth ($b > 1$), the curve starts near a horizontal asymptote (usually the $x$-axis), rises slowly, then steeply, never touching the asymptote on the low side. For decay ($0 < b < 1$), it falls from high on the left toward the asymptote on the right. The $y$-intercept is always the initial value $a$ (the output at $x = 0$), since $b^{0} = 1$. Unlike a line, an exponential curve bends, and its rate of change itself grows or shrinks, which is why exponential growth ultimately outpaces any straight line.

Doubling and half-life

Two common phrasings are doubling time and half-life. A quantity that doubles every period is $a \cdot 2^{t}$; one that halves every period is $a \cdot \left(\frac{1}{2}\right)^{t}$. If a sample halves every 5 years, after $t$ years the fraction remaining is $\left(\frac{1}{2}\right)^{t/5}$, so the exponent is the number of half-lives elapsed. Reading "doubles every 3 hours" as a base of 2 with the exponent counting 3-hour periods turns these word problems into a direct substitution.

Why the growth pattern matters

Recognising that a quantity multiplies (rather than adds) each step tells you to use an exponential model and to read the base as a percent change. The most reliable habit is to translate the factor immediately: $1.04$ means $+4\%$, $0.90$ means $-10\%$. For logs, rewrite in exponential form. These two translations cover nearly every exponential and logarithmic question the ACT asks.

Try this

Q1. In $V(t) = 500(0.8)^{t}$, what is the decay rate per period? [1 point]

• Cue. $b = 0.8 = 1 - 0.2$, so a 20% decrease per period.

Q2. Evaluate $\log_3 81$. [1 point]

• Cue. $3^{x} = 81 = 3^{4}$, so $\log_3 81 = 4$.