Mathematics · Functions

Exponential Growth and Decay Functions

Reference entry · last updated September 20, 2026

An exponential function has the form \( f(x) = a \cdot b^x \), where \( a \neq 0 \), \( b > 0 \), and \( b \neq 1 \). The quantity changes by a constant ratio per unit of input rather than by a constant amount, which is what separates it from a linear function.

1. First principles

With \( a > 0 \), the domain is all real numbers and the range is \( (0, \infty) \). The initial value is \( f(0) = a \), and each unit step multiplies the previous value by \(b\) [1, 2]. A linear function has a constant difference between successive equal steps; an exponential function has a constant ratio.

2. Growth factor and growth rate

The base \(b\) is the growth factor. The growth rate \(r\) is the fractional change per period:

\( b = 1 + r, \qquad f(x) = a(1 + r)^x \)

When \( b > 1 \) the function grows and \( r > 0 \). When \( 0 < b < 1 \) the function decays and \( r < 0 \). A growth factor of \(1.05\) is a growth rate of \(5\%\); a decay factor of \(0.8\) is a rate of \(-20\%\).

3. Continuous form and the number e

Compounding can be made continuous by letting the number of compounding periods grow without bound. The limit defines the constant \( e \approx 2.71828\):

\( e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \)

The continuous form of an exponential function is \( f(t) = a e^{kt} \), where \(k\) is the continuous growth rate. The two forms are equivalent with \( b = e^{k} \), so \( k = \ln b \).

4. Doubling time and half-life

Doubling time is the input needed for the value to double. Setting \( a b^t = 2a \) and solving:

\( t_{\text{double}} = \frac{\ln 2}{\ln b} \)

Half-life is the input needed for the value to halve, which is the same computation with \(\ln 2\) in the numerator and a base below \(1\):

\( t_{\text{half}} = \frac{\ln(1/2)}{\ln b} = \frac{\ln 2}{\ln(1/b)} \)

Both periods are constant and independent of the starting value.

5. Compound interest

An amount \(P\) at annual rate \(r\) compounded \(n\) times per year for \(t\) years grows to:

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

As \(n \to \infty\) this approaches the continuous-compounding limit \( A = P e^{rt} \). Compound interest is the canonical financial application of exponential growth.

6. Graph features

The graph passes through \((0, a)\) and \((1, ab)\).
The horizontal line \(y = 0\) is a horizontal asymptote; the graph approaches it without touching.
For \(b > 1\) with \(a > 0\), the curve rises increasingly steeply to the right.
Transformations apply as for any function: vertical shift moves the asymptote, and reflection gives \(f(x) = a b^{-x}\).
If \(b = 1\), the function is constant and not exponential.

References

^OpenStax, College Algebra 2e, Rice University (Exponential functions).
^James Stewart, Lothar Redlin, and Saleem Watson, Precalculus: Mathematics for Calculus, 7th ed., Cengage Learning, 2015.