An exponential function has the form f(x) = ab^x, where a is the initial value of the function when x = 0 and b > 0 is the multiplier. For values of b between 0 and 1, the function represents exponential decay (the value of the function decreases as x increases). For values of b greater than 1, the function represents exponential growth (the value of the function increases as x increases). If b = 1, the function simplifies to the constant function f(x) = a.

A common application of exponential functions is measuring a quantity over time that is increasing or decreasing by a fixed rate each time period. Expressing that rate as a decimal (e.g., r = 15% = 0.15), we write b = 1+r, so that the exponential function becomes f(x) = a(1+r)^x.

Example 1

• Find the equation for an exponential function representing annual sales if sales at the beginning of the time period in question is 10,000 and sales increases by 5% each year.
• Since initial sales are 10,000, a = 10,000.
• Since the growth rate is 5%, r = 0.05, and b = 1+r = 1.05.
• The sales function is f(x) = 10,000(1.05)^x.
• Check: after 1 year, sales are 10,000(1.05)¹ = 10,500.
• Check: after 2 years, sales are 10,000(1.05)² = 11,025.
• Check: after 10 years, sales are 10,000(1.05)¹⁰ = 16,289.

Example 2

• Find the equation for an exponential function representing daily radioactive decay of an isotope if the mass at the beginning of the time period in question is 1,000 mg and mass decreases by 12% each day.
• Since the initial mass is 1,000, a = 1,000.
• Since the decay rate is 12%, r = –0.12, and b = 1+r = 0.88.
• The decay function is f(x) = 1,000(0.88)^x.
• Check: after 1 day, the mass is 1,000(0.88)¹ = 880 mg.
• Check: after 2 day, the mass is 1,000(0.88)² = 774.4 mg.
• Check: after 1 week, the mass is 1,000(0.88)⁷ = 408.7 mg.

Find an exponential equation given an initial value and a later value

Example 3

• Find the equation for an exponential function representing annual growth of an animal population if the population was 120 in 2019 and 230 in 2024.
• Since the initial population is 120, a = 120.
• Since the population is 230 after 5 years, we can write 120b⁵ = 230.
• Solving for b gives: [latex]b=\sqrt[5]{\frac{230}{120}}=\left(\frac{230}{120}\right)^{1/5}=1.1390[/latex].
• The population function is f(x) = 120(1.1390)^x.
• Check: after 5 years, the population is 120(1.1390)⁵ = 230.00.

Find an exponential equation given two arbitrary values

Example 4

• Find the equation for an exponential function if the value of the function at x = 3 is f(x) = 5 and the value of the function at x = 7 is f(x) = 24.
• The equation based on the first value is ab³ = 5.
• Solving for a gives: a = 5/b³.
• The equation based on the second value is ab⁷ = 24.
• Substitute a = 5/b³ into the second equation: 5b⁷/b³ = 5b⁴ = 24.
• Solving for b gives: [latex]b=\sqrt[4]{\frac{24}{5}}=\left(\frac{24}{5}\right)^{1/4}=1.4802[/latex].
• Solving for a gives: a = 5 / 1.4802³ = 1.5417.
• The exponential function is f(x) = 1.5417(1.4802)^x.
• Check: f(3) = 1.5417(1.4802)³ = 5.00.
• Check: f(7) = 1.5417(1.4802)⁷ = 24.00.