Exponential Functions & Growth

Linear functions grow by a constant amount each step. Exponential functions grow by a constant factor each step -- and that single difference changes everything. Populations explode, investments compound, and radioactive samples vanish, all following the same elegant pattern.

The Exponential Model

An exponential function has the form:

y = a \cdot b x

Symbol	Meaning
a	Initial value (y-intercept, when x = 0)
b	Base -- the constant multiplier each step
x	Exponent (often time)

Growth vs. Decay

Growth (b > 1)

Each step, the quantity multiplies by more than 1, so it increases.

Example: b = 1.05 means 5% growth per period.

Decay (0 < b < 1)

Each step, the quantity multiplies by less than 1, so it shrinks.

Example: b = 0.92 means 8% decay per period.

Connecting Rate to Base

If the growth rate is r (as a decimal), then b = 1 + r for growth and b = 1 - r for decay. A 7% annual increase means b = 1.07. A 3% annual decrease means b = 0.97.

The Number e

The irrational constant e ≈ 2.71828... appears naturally when compounding happens continuously. If you invest $1 at 100% interest compounded n times per year, the amount approaches e as n grows without bound:

e = lim (1 + 1/n) n \approx 2.71828

Continuous growth uses the model y = a · e^kt, where k is the continuous rate.

Compound Growth

The compound interest formula is a direct application:

A = P(1 + r/n) nt

P = principal, r = annual rate, n = compounding periods per year, t = years, A = final amount.

Worked Example 1 -- Investment

You invest $2,000 at 6% annual interest compounded monthly. How much after 5 years?

Identify values: P = 2000, r = 0.06, n = 12, t = 5.
Substitute: A = 2000(1 + 0.06/12)^12·5 = 2000(1.005)⁶⁰.
Compute: (1.005)⁶⁰ ≈ 1.34885. So A ≈ 2000 × 1.34885 ≈ $2,697.70.

Worked Example 2 -- Population

A town of 50,000 grows at 2% per year. Write a model and predict the population in 10 years.

Model: P(t) = 50000 · 1.02^t.
At t = 10: P(10) = 50000 · 1.02¹⁰ ≈ 50000 × 1.21899 ≈ 60,950.

Worked Example 3 -- Radioactive Decay

A 200 g sample decays at 4% per hour. How much remains after 6 hours?

Model: A(t) = 200 · 0.96^t (since decay of 4% means b = 1 - 0.04 = 0.96).
A(6) = 200 · 0.96⁶ ≈ 200 × 0.7828 ≈ 156.6 g.

Common Mistake

Do not confuse the rate with the base. A 5% growth rate does not mean b = 5 or b = 0.05. It means b = 1.05. Always ask: "What do I multiply by each period?"

Practice Problems

Write an exponential model for a bacteria colony that starts at 300 and doubles every hour.
Show Solution

Doubling means b = 2. Model: N(t) = 300 · 2^t, where t is in hours.
A car worth $25,000 depreciates 15% per year. What is it worth after 3 years?
Show Solution

b = 1 - 0.15 = 0.85. V(3) = 25000 · 0.85³ = 25000 × 0.614125 ≈ $15,353.13.
$5,000 is invested at 4% annual interest compounded quarterly. Find the amount after 10 years.
Show Solution

A = 5000(1 + 0.04/4)^4·10 = 5000(1.01)⁴⁰ ≈ 5000 × 1.48886 ≈ $7,444.32.
The function f(x) = 800 · (0.75)^x models a quantity. Is this growth or decay? What percentage change occurs each step?
Show Solution

Since b = 0.75 < 1, this is decay. The quantity decreases by 25% each step (1 - 0.75 = 0.25).
A population grows continuously at rate k = 0.03 per year from an initial size of 10,000. Find the population after 20 years using P = 10000 · e^0.03·20.
Show Solution

P = 10000 · e^0.6 ≈ 10000 × 1.8221 ≈ 18,221.

Summary

Exponential functions have the form y = a · b^x.
b > 1 gives growth; 0 < b < 1 gives decay.
The number e ≈ 2.718 arises from continuous compounding.
Compound interest formula: A = P(1 + r/n)^nt.