Difficulty: Intermediate | Prerequisites: Properties of logarithms, basic algebra with e and ln
Exponential functions model anything that grows or shrinks at a rate proportional to its current size: bank accounts earning interest, radioactive isotopes decaying, drug concentrations fading. This section teaches you to plug into the compound interest formula, switch between discrete and continuous compounding, solve for any unknown variable using logarithms, and apply the same machinery to half-life and doubling-time problems. You need to be comfortable with properties of ln and e before diving in.
Money grows via A = P(1 + r/n)^(nt) for discrete compounding and A = Pe^(rt) for continuous compounding. Radioactive decay and similar processes use the same continuous model with a negative rate. Solving for time, rate or principal means isolating the variable with ln. Half-life is the time for a quantity to halve; doubling time is the time for it to double.
Compound interest (discrete)
Interest calculated on both the initial principal and the accumulated interest from previous periods. The compounding frequency (annually, monthly, daily) determines how often interest is added.
In simple terms, your interest earns interest.
Continuous compounding
The mathematical limit of compounding as the number of compounding periods per year approaches infinity. Uses the formula A = Pe^(rt). This is the model used most often in calculus courses because it pairs naturally with derivatives.
Principal (P)
The initial amount invested or deposited.
Future value (A)
The amount the investment is worth after t years of growth.
Present value
The amount you would need to invest now in order to reach a specified future value. Found by solving the compound interest formula for P.
Think of it as "what is that future payout worth in today's money?"
Annual interest rate (r)
The yearly rate of growth, expressed as a decimal (10% = 0.10).
Half-life
The time it takes for a quantity to reduce to half its initial value. Used in radioactive decay and similar processes. Calculated by setting A = P/2 and solving for t.
Doubling time
The time it takes for a quantity to double. Calculated by setting A = 2P and solving for t.
Natural logarithm, ln(x)
The logarithm base e. Used to "undo" the exponential when solving for a variable in the exponent. Key property: ln(e^a) = a.
Formula: A = P(1 + r/n)^(nt)
P = principal, r = annual rate (decimal), n = compounding periods per year, t = years
Annually: n = 1. Monthly: n = 12. Daily: n = 365.
Example: $5,000 at 10% for 8 years.
Annually: A = 5000(1 + 0.10)^8 = $10,717.94
Monthly: A = 5000(1 + 0.10/12)^(96) = $11,090.88
Daily: A = 5000(1 + 0.10/365)^(2920) = $11,126.49
Formula: A = Pe^(rt)
Same example continuously: A = 5000e^(0.80) = $11,127.70
Continuous compounding always yields slightly more than daily, but the difference is small.
Rearrange A = Pe^(rt) to get P = Ae^(-rt).
Example: a note pays $20,000 in 10 years at 5.2% continuous. P = 20000e^(-0.052 * 10) = 20000e^(-0.52) = $11,890.41.
Set up the equation, divide both sides by P, then take ln of both sides.
Doubling time at 4% continuous: 2P = Pe^(0.04t), so 2 = e^(0.04t), ln 2 = 0.04t, t = ln 2 / 0.04 = 17.33 years.
Same approach: isolate the exponential, take ln, solve for r.
Money doubles in 13 years continuously: 2 = e^(13r), ln 2 = 13r, r = ln 2 / 13 = 0.0533, so 5.33%.
Jake's $10,000 grows to $12,000 in 5 years: 12000 = 10000e^(5r), 1.2 = e^(5r), ln 1.2 = 5r, r = ln 1.2 / 5 = 0.0365, so 3.65%.
To solve 4 = 3e^(0.01x): divide by 3 to get 4/3 = e^(0.01x), take ln of both sides: ln(4/3) = 0.01x, so x = ln(4/3) / 0.01 = 28.77.
Uses A = Pe^(rt) with r negative.
Finding the decay rate from half-life: set A = P/2, so 1/2 = e^(rt), ln(1/2) = rt, r = ln(1/2) / t = -ln 2 / t.
Strontium, half-life 90 years: r = -ln 2 / 90 = -0.0077, or -0.77%.
Finding half-life from decay rate: t = -ln 2 / r.
r = -0.003 per year: t = -ln 2 / (-0.003) = 231 years.
Time to decay to a fraction: set A/P = desired fraction, solve.
Radium, half-life 1690 years, 50g to 5g: 5/50 = e^(rt) where r = -ln 2 / 1690. So 0.1 = e^(rt), ln 0.1 = rt, t = ln 0.1 / r = ln 0.1 / (-ln 2 / 1690) = 5614 years.
A = P\left(1 + \frac{r}{n}\right)^{nt} \quad \text{(discrete compounding)}A = Pe^{rt} \quad \text{(continuous compounding)}P = Ae^{-rt} \quad \text{(present value, continuous)}t_{\text{double}} = \frac{\ln 2}{r} \quad \text{(doubling time, continuous)}t_{\text{half}} = \frac{\ln 2}{|r|} \quad \text{(half-life from decay rate)}r = \frac{-\ln 2}{t_{\text{half}}} \quad \text{(decay rate from half-life)}Every savings account, mortgage and bond valuation uses compound interest. Continuous compounding is the standard model in mathematical finance. Half-life calculations are used in nuclear medicine (how long until a tracer leaves the body), archaeology (carbon-14 dating) and pharmacology (drug clearance rates).
Forgetting to convert the percentage to a decimal. 10% is 0.10, not 10. This is the single most common arithmetic error in these problems.
Using the wrong n. Monthly means n = 12, not n = 1. Daily means n = 365. Continuously means you switch to A = Pe^(rt) entirely.
Taking ln too early or too late. You must isolate the exponential expression first (divide both sides by P or by the coefficient), then take ln. Taking ln of a sum does not simplify: ln(A + B) is not ln A + ln B.
Sign errors in decay problems. The decay rate r is negative. When the problem gives a half-life, compute r = -ln 2 / t (negative). When it gives a negative r and asks for half-life, use t = ln 2 / |r|.
Confusing "compounded continuously" with "compounded annually." Different formulas, different answers.
⚠️ Expect at least one problem where you compute A under multiple compounding frequencies (annually, monthly, daily, continuously) for the same principal, rate and time.
⚠️ Present value problems ("how much should you pay now?") require you to solve for P, not A.
⚠️ Doubling time and half-life questions test whether you can set up the equation and use ln correctly.
⚠️ The algebra-review problem (solving 4 = 3e^(0.01x)) tests the same ln technique without a word-problem wrapper. Do not overthink it.
True or False: Continuous compounding always gives a larger result than daily compounding for the same P, r and t. (True.)
Fill in the blank: To find present value under continuous compounding, use P = ___. (Ae^(-rt))
True or False: ln(e^5) = 5. (True.)
Fill in the blank: The doubling time for an investment at rate r compounded continuously is t = ___. (ln 2 / r)
True or False: A half-life of 90 years means the decay rate is r = -90. (False. r = -ln 2 / 90 = -0.0077.)
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 = 0.2877 / 0.01 = 28.77.
Q: $5,000 is invested at 10% for 8 years, compounded monthly. What is the future value?
A: A = 5000(1 + 0.10/12)^(12 * 8) = 5000(1.00833...)^96 = $11,090.88.
Q: A note pays $20,000 in 10 years. Money is worth 5.2% compounded continuously. What should you pay now?
A: P = 20000e^(-0.052 * 10) = 20000e^(-0.52) = $11,890.41.
Q: How long will it take an investment to double at 4% compounded continuously?
A: 2 = e^(0.04t), ln 2 = 0.04t, t = 0.6931 / 0.04 = 17.33 years.
Q: A bank doubles your money in 13 years with continuous compounding. What is the annual rate?
A: 2 = e^(13r), ln 2 = 13r, r = 0.6931 / 13 = 0.0533, so 5.33%.
Q: A strontium isotope has a half-life of 90 years. Find the continuous decay rate.
A: r = -ln 2 / 90 = -0.6931 / 90 = -0.0077, or -0.77%.
Q: A substance has a decay rate r = -0.3% per year. What is its half-life?
A: t = ln 2 / 0.003 = 0.6931 / 0.003 = 231 years.
Q: Radium has a half-life of 1,690 years. How long until 50g decays to 5g?
A: r = -ln 2 / 1690. Set 5 = 50e^(rt), so 0.1 = e^(rt), ln 0.1 = rt. t = ln 0.1 / (-ln 2 / 1690) = (-2.3026) / (-0.000410) = 5,614 years.
Continuous compounding leads directly into differentiation of exponential functions (the next topic on the exam). The derivative of A = Pe^(rt) with respect to t is A' = Pr * e^(rt), which gives the instantaneous rate of change of the investment. You will see this exact calculation in problems 24 and 25 of the review.
The ln properties tested here (expanding, simplifying) also show up in differentiation problems 17-21, where you must rewrite expressions using log rules before taking derivatives.
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