Difficulty: Intermediate | Prerequisites: Algebra (logarithms, exponent rules)
Exponential functions model any quantity that grows or shrinks at a rate proportional to its current size. In a business calculus course, this shows up everywhere: compound interest on investments, depreciation of assets, and population or price growth. You need to be comfortable with logarithms and basic exponent rules before this material will click. If Chapter 3 (limits and basic derivatives) is shaky, shore that up first.
Money and other quantities that grow or shrink by a fixed percentage follow exponential models. The two main formulas are B = Pe^(rt) for continuous compounding and B = P(1 + r/n)^(nt) for periodic compounding. Most exam problems ask you to plug in what you know, solve for the one unknown (usually time or the starting amount), and use ln to undo the exponential when the unknown is in the exponent.
Continuous compounding (B = Pe^(rt))
The balance when interest is applied every instant, not at fixed intervals. "e" (Euler's number, approximately 2.718) is the base. In simple terms, this is the theoretical maximum growth you get when compounding happens infinitely often.
Discrete (periodic) compounding (B = P(1 + r/n)^(nt))
The balance when interest is added at regular intervals: quarterly (n = 4), semiannually (n = 2), monthly (n = 12), etc. Think of it as: "every n periods per year, the bank adds a fraction of the annual rate to your balance."
Principal (P)
The starting amount of money invested or owed. In simple terms, this is how much you put in at the beginning.
Growth rate (r > 0)
The rate at which the quantity increases per unit time, expressed as a decimal. A 5% annual rate means r = 0.05. If r is positive, the quantity grows.
Decay rate (r < 0)
The same formula applies when a quantity shrinks over time (depreciation, radioactive decay). A negative r means the exponential curve slopes downward. Think of it as: growth in reverse.
Natural logarithm (ln)
The inverse of e^x. Used to "undo" the exponential when you need to solve for time or rate. ln(e^x) = x, and e^(ln x) = x. In simple terms, ln answers the question: "e raised to what power gives me this number?"
Euler's number (e)
The irrational constant approximately equal to 2.71828. It is the base of the natural logarithm and appears naturally whenever growth is proportional to current size.
Use this formula when a problem says "compounded continuously" or when you are modelling exponential growth/decay without a stated compounding interval.
B = the future (ending) amount
P = the present (starting) amount
r = the continuous growth or decay rate (decimal)
t = time
Finding the rate when you have two data points
Plug in P, B, and t, then isolate r:
Divide both sides by P to get B/P = e^(rt)
Take ln of both sides: ln(B/P) = rt
Divide by t: r = ln(B/P) / t
Worked Example 1: Exponential rent increase
Rent was $800 and four months later it was $1,200. If it grows exponentially, what will it be in a year (12 months)?
Step 1: Find r. 1200 = 800 e^(r * 4). Divide: 3/2 = e^(4r). Take ln: ln(3/2) = 4r. So r = ln(3/2) / 4.
Step 2: Find B at t = 12. B = 800 e^(r * 12) = 800 e^(12 * ln(3/2)/4) = 800 e^(3 ln(3/2)) = 800 (3/2)^3 = 800 * 27/8 = $2,700.
Shortcut insight: 12 months = 3 intervals of 4 months. Each interval multiplies by 3/2. So 800 * (3/2)^3 = $2,700.
Worked Example 7 (from the review): Continuous bond growth
Kobe invests $10,000 at 5% compounded continuously. How long to reach $15,000?
15,000 = 10,000 e^(0.05t)
3/2 = e^(0.05t)
ln(3/2) = 0.05t
t = ln(3/2) / 0.05
Quiz Example 1: $4,000 to $10,000 at 3% continuous
10,000 = 4,000 e^(0.03t)
5/2 = e^(0.03t)
t = ln(5/2) / 0.03
Use this when the problem states a compounding frequency: quarterly, semiannually, monthly, etc.
n = number of compounding periods per year
r = annual rate (decimal)
t = time in years
Worked Example 2: Quarterly investment
Invest $10,000 at 5% compounded quarterly for 4 years.
B = 10,000 (1 + 0.05/4)^(4*4) = 10,000 (1 + 0.0125)^16
Calculate (1.0125)^16 to find B.
Quiz Example 2: Saving for FIU law school tuition
Tuition will be $100,000 in 5 years. Interest is 5% compounded semiannually. How much to invest now?
100,000 = P (1 + 0.05/2)^(2*5) = P (1.025)^10
(1.025)^10 = 1.28 (approx.)
P = 100,000 / 1.28 = $78,125
This is solving for the present value: rearrange the formula to P = B / (1 + r/n)^(nt).
Same continuous model, but the quantity decreases. The rate r will be negative.
Quiz Example 3: Machine depreciation
A machine cost $30,000 and was worth $20,000 after 2 years. What is it worth after 8 years?
Step 1: Find r. 20,000 = 30,000 e^(2r). So 2/3 = e^(2r). r = ln(2/3) / 2.
Step 2: Find B at t = 8. B = 30,000 e^(r * 8). Since r = ln(2/3)/2, this simplifies to B = 30,000 e^(4 ln(2/3)) = 30,000 (2/3)^4.
Note that ln(2/3) is negative, confirming the value is declining.
B = P e^{rt}Continuous compounding. Use when no compounding interval is stated, or the problem says "continuously."
B = P\left(1 + \frac{r}{n}\right)^{nt}Discrete compounding. n = number of compounding periods per year.
r = \frac{\ln(B/P)}{t}Solving for the continuous growth/decay rate given two data points.
t = \frac{\ln(B/P)}{r}Solving for time in a continuous compounding problem.
P = \frac{B}{\left(1 + \frac{r}{n}\right)^{nt}}Solving for the present value (how much to invest now) in a discrete compounding problem.
Key log identities used in these problems:
ln(e^x) = x
e^(ln x) = x
ln(1) = 0
ln(a/b) = ln(a) - ln(b)
Continuous compounding models how banks price certain financial products, and it is the standard model in economics for inflation-adjusted returns. Exponential decay is how accountants model asset depreciation and how scientists describe radioactive half-lives.
Students often confuse the two formulas. If the problem says "continuously," use B = Pe^(rt). If it names a frequency (quarterly, semiannually, monthly), use B = P(1 + r/n)^(nt). Mixing them up is the single most common error on exams.
Students forget to convert the percentage to a decimal. 5% means r = 0.05, not r = 5.
When solving for t, students sometimes try to divide both sides by e instead of taking ln. You cannot divide by e to remove an exponent. You take the natural log.
Students assume that doubling the rate doubles the final amount. It does not; the relationship is exponential, not linear.
Expect at least one problem that gives you two data points (a starting value and a later value) and asks you to find r, then project forward to a new time. This is the two-step pattern from Examples 1, 3, and 7.
Expect a present-value problem: "How much should you invest now to have $X in Y years?" This means solving for P.
The ln step is where most algebra mistakes happen. Practise isolating the exponent cleanly.
Know that ln(1) = 0. It simplifies many expressions and is easy to forget under pressure.
True or False: B = Pe^(rt) is used when interest compounds quarterly. (False: use B = P(1 + r/n)^(nt) with n = 4.)
Fill in the blank: To solve for t in a continuous compounding problem, take the ______ of both sides after isolating the exponential. (natural logarithm / ln)
True or False: If r is negative in B = Pe^(rt), the quantity is growing. (False: a negative r means decay.)
Fill in the blank: ln(e^x) = ______. (x)
True or False: Compounding quarterly at 5% gives the same result as compounding continuously at 5%. (False: continuous compounding yields slightly more.)
Q: You invest $6,000 in an account that compounds continuously at 4%. How long until the investment reaches $9,000?
A: 9,000 = 6,000 e^(0.04t). Divide: 3/2 = e^(0.04t). Take ln: ln(3/2) = 0.04t. So t = ln(3/2) / 0.04 years.
Q: A car depreciates exponentially from $25,000 to $18,000 in 3 years. What is its continuous depreciation rate?
A: 18,000 = 25,000 e^(3r). Divide: 18/25 = e^(3r). Take ln: ln(18/25) = 3r. So r = ln(18/25) / 3. Since 18/25 < 1, ln is negative, confirming decay.
Q: You need $50,000 in 6 years. An account pays 6% compounded semiannually. How much should you invest today?
A: 50,000 = P(1 + 0.03)^12. P = 50,000 / (1.03)^12.
Q: A population doubles every 5 years under continuous growth. What is the growth rate r?
A: 2P = P e^(5r). So 2 = e^(5r). r = ln(2) / 5.
Q: $8,000 is invested at 5% compounded quarterly for 10 years. Write the expression for the final balance.
A: B = 8,000 (1 + 0.05/4)^(4 * 10) = 8,000 (1.0125)^40.
This material connects directly to derivatives and antiderivatives: the derivative of e^(rt) is re^(rt), which is why exponential functions model proportional growth. In Module 4 you will also see definite integrals used to measure accumulated change (population increase, total depreciation), which is the flip side of the rate-of-change story here.
Exponential growth, exponential decay, compound interest, continuous compounding, discrete compounding, Euler's number, natural logarithm, ln, e, present value, future value, growth rate, decay rate, B = Pe^rt, B = P(1+r/n)^nt, time value of money, FIU Calculus for Business, Module 4, Professor Fox, depreciation, semiannual compounding, quarterly compounding