Difficulty: Intermediate | Prerequisites: Properties of exponents and logarithms, basic algebra, familiarity with the number e
Exponential functions model anything that grows or decays at a rate proportional to its current value: bank accounts earning compound interest, radioactive isotopes losing mass, drug concentrations fading in the bloodstream. Section 3.2 of MAC 2233 applies these models to finance (compound interest, present value) and science (half-life, continuous decay). You need to be comfortable with the natural logarithm, ln, as the tool for "undoing" exponentials when solving for time or rate. If logarithm properties feel unfamiliar, revisit those before working through these problems.
Money in a bank account grows according to A = P(1 + r/n)^(nt) for periodic compounding, or A = Pe^(rt) for continuous compounding. Radioactive substances decay the same way, with a negative rate. To solve for time or rate, take ln of both sides and isolate the variable.
Compound interest
Interest calculated on the initial principal and also on the accumulated interest from previous periods.
Think of it as interest earning interest: the balance snowballs over time.
Continuous compounding
The mathematical limit of compound interest when the number of compounding periods per year approaches infinity. Uses the formula A = Pe^(rt).
In simple terms, the bank calculates interest every instant rather than monthly or daily.
Present value
The amount of money you would need to invest today to reach a desired future amount, given a particular interest rate and time period.
Think of it as working the compound interest formula backwards: you know A and solve for P.
Half-life
The time it takes for a quantity undergoing exponential decay to reduce to half its initial value.
In simple terms, if a substance has a half-life of 90 years, every 90 years you have half as much.
Continuous decay rate (r)
The constant in A = Pe^(rt) when the quantity is shrinking. The rate r is negative for decay.
Natural logarithm, ln(x)
The logarithm with base e. It is the inverse of the exponential function e^x.
Think of ln as the tool you use to "undo" an exponential when solving for an unknown exponent.
e (Euler's number)
The mathematical constant approximately equal to 2.71828. It is the base of continuous exponential growth and decay.
Formula: A = P(1 + r/n)^(nt)
A = future amount
P = principal (initial investment)
r = annual interest rate (as a decimal)
n = number of compounding periods per year (annually = 1, monthly = 12, daily = 365)
t = time in years
Example: $5,000 at 10% for 8 years, compounded annually: A = 5000(1 + 0.1/1)^(1·8) = 5000(1.1)^8 ≈ $10,717.94
Compounded monthly: A = 5000(1 + 0.1/12)^(12·8) ≈ $11,098.20
Compounded daily: A = 5000(1 + 0.1/365)^(365·8) ≈ $11,126.50
Formula: A = Pe^(rt)
This is the limit of periodic compounding as n approaches infinity
Example: $5,000 at 10% for 8 years, compounded continuously: A = 5000e^(0.1·8) = 5000e^(0.8) ≈ $11,127.70
Continuous compounding always yields slightly more than daily compounding, but the difference is small
To find how long it takes to reach a target amount, isolate the exponential and take ln of both sides
Example: how long to double at 4% continuous? 2P = Pe^(0.04t), so 2 = e^(0.04t), ln(2) = 0.04t, t = ln(2)/0.04 ≈ 17.3 years
Same approach: isolate e^(rt), take ln, solve for r
Example: $10,000 grows to $12,000 in 5 years continuously. 12000 = 10000e^(5r), so 1.2 = e^(5r), ln(1.2) = 5r, r = ln(1.2)/5 ≈ 0.0365 or 3.65%
Rearrange A = Pe^(rt) to get P = Ae^(−rt)
Example: to have $20,000 in 10 years at 5.2% continuous, invest P = 20000e^(−0.052·10) = 20000e^(−0.52) ≈ $11,890
Decay uses the same formula with a negative r: A = Pe^(rt) where r < 0
Half-life: the time t when A = P/2, so (1/2) = e^(rt), giving t = ln(1/2)/r = −ln(2)/r
Example: strontium with a 90-year half-life. (1/2)P = Pe^(90r), so r = ln(1/2)/90 ≈ −0.0077
Reverse problem: given r = −0.3% per year, find half-life. t = ln(1/2)/(−0.003) ≈ 231 years
To solve 4 = 3e^(0.01x): divide both sides by 3 to get 4/3 = e^(0.01x), then ln(4/3) = 0.01x, so x = ln(4/3)/0.01 ≈ 28.77
A = P\left(1 + \frac{r}{n}\right)^{nt} \quad \text{(periodic compounding)}A = Pe^{rt} \quad \text{(continuous compounding)}P = Ae^{-rt} \quad \text{(present value, continuous)}t = \frac{\ln(A/P)}{r} \quad \text{(solving for time, continuous)}r = \frac{\ln(A/P)}{t} \quad \text{(solving for rate, continuous)}t_{1/2} = \frac{-\ln 2}{r} = \frac{\ln 2}{|r|} \quad \text{(half-life from decay rate)}r = \frac{\ln(1/2)}{t_{1/2}} \quad \text{(decay rate from half-life)}Banks and investment firms use continuous compounding models to price bonds and value annuities. The present value formula is the basis of discounted cash flow analysis, a standard method for valuing companies and projects. Half-life calculations appear in pharmacology (how quickly a drug clears the body), nuclear physics (dating fossils with carbon-14) and environmental science (tracking pollutant decay).
Students often forget to convert percentage rates to decimals before plugging into formulas. 10% is r = 0.10, not r = 10.
When solving for t or r, students sometimes try to "bring down" the exponent without using ln. You cannot simplify e^(0.04t) = 2 by dividing both sides by e. You must take ln of both sides.
Half-life problems use a negative r. Students sometimes write r as positive and then get a negative half-life, which is nonsensical. Keep r negative for decay, or use the absolute value formula t = ln(2)/|r|.
"Compounded continuously" and "compounded daily" are different. Continuous uses A = Pe^(rt). Daily uses A = P(1 + r/365)^(365t). The answers will be close but not identical.
⚠️ Expect at least one problem requiring you to solve for t using ln. Know the steps: isolate the exponential, take ln of both sides, solve algebraically.
⚠️ Present value problems ("how much should you pay now?") are a favourite. Recognise the backwards setup: you know A and solve for P.
⚠️ You may be asked to compute the same investment under annual, monthly, daily and continuous compounding. Have all four formulas ready and know which n value goes with each.
⚠️ Half-life and doubling-time problems follow the same algebraic pattern. If you can solve one, you can solve the other.
Fill in the blank: For continuous compounding, the formula is A = P · e^(______). (rt)
True or False: To find the half-life, you set A = 2P in the decay formula. (False, you set A = P/2.)
Fill in the blank: To solve for t in e^(0.04t) = 2, the first step is to take ______ of both sides. (ln)
True or False: Compounding monthly gives a higher return than compounding continuously. (False, continuous compounding always yields the most.)
Fill in the blank: Present value is found by rearranging A = Pe^(rt) to P = ______. (Ae^(−rt))
Q: Solve for x: 4 = 3e^(0.01x)
A: Divide both sides by 3: 4/3 = e^(0.01x). Take ln: ln(4/3) = 0.01x. So x = ln(4/3)/0.01 ≈ 28.77.
Q: $5,000 is invested at 10% annual interest for 8 years. Find the value if compounded annually, monthly, daily and continuously.
A: Annually: A = 5000(1.1)^8 ≈ $10,717.94. Monthly: A = 5000(1 + 0.1/12)^(96) ≈ $11,098.20. Daily: A = 5000(1 + 0.1/365)^(2920) ≈ $11,126.50. Continuously: A = 5000e^(0.8) ≈ $11,127.70.
Q: A note will pay $20,000 at maturity in 10 years. How much should you pay now if money is worth 5.2% compounded continuously?
A: P = 20000e^(−0.052 × 10) = 20000e^(−0.52) ≈ $11,890.
Q: How long will it take an investment to double at 4% compounded continuously?
A: 2P = Pe^(0.04t), so 2 = e^(0.04t), ln(2) = 0.04t, t = ln(2)/0.04 ≈ 17.3 years.
Q: Money doubles every 13 years with continuous compounding. What is the annual interest rate?
A: 2 = e^(13r), so ln(2) = 13r, r = ln(2)/13 ≈ 0.053 or 5.3%.
Q: A strontium isotope has a half-life of 90 years. What is the continuous compound rate of decay?
A: (1/2)P = Pe^(90r), so ln(1/2) = 90r, r = ln(1/2)/90 ≈ −0.0077 or −0.77%.
Q: The half-life of radium is 1,690 years. How long for a 50g sample to reduce to 5g?
A: First find r: r = ln(1/2)/1690 ≈ −0.000410. Then solve 5 = 50e^(rt): 0.1 = e^(rt), ln(0.1) = rt, t = ln(0.1)/r = (−2.3026)/(−0.000410) ≈ 5,616 years.
The continuous compounding formula A = Pe^(rt) reappears when you differentiate exponential functions in Section 3.3: the derivative of e^(rt) with respect to t is re^(rt), which is the instantaneous rate of change of the account balance. Marginal analysis problems (Section 2.7) sometimes use exponential cost or revenue functions, combining these two topic areas. Later in the course, optimisation problems may ask you to maximise continuously compounded investment returns.
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