Unit 2: Exponential and Logarithmic Functions

Unit 2 covers exponential and logarithmic functions: how to tell exponential growth apart from linear growth, rewriting equivalent exponential forms, compound and continuous growth, the definition and properties of logarithms, solving exponential and logarithmic equations, semi-log plots, and logistic models.

2.1 Exponential vs Linear Growth

A linear function grows by constant difference. Each step forward adds the same amount: f(x) = mx + c has a slope of m, so every unit increase in x raises the output by m. An exponential function grows by constant percent change. Each step forward multiplies the output by the same factor: f(x) = a · b^x has a growth factor b, so every unit increase in x multiplies the output by b. Growth means b > 1; decay means 0 < b < 1.

The table below compares E(x) = 20(1.1)^x, which grows 10% per step, with L(x) = 4x + 20, which adds 4 per step. For the exponential column, each entry is 1.1 times the one before: 20 × 1.1 = 22, 22 × 1.1 = 24.2, 24.2 × 1.1 = 26.62, 26.62 × 1.1 = 29.282. For the linear column, each entry is 4 more than the one before. Constant ratio marks exponential growth; constant difference marks linear growth.

2.2 Equivalent Forms of Exponential Functions

The same growth can be written over different time units, and the rates convert through the equivalence (1 + r_annual) = (1 + r_monthly)¹². If a quantity grows 7% per year, the monthly growth factor r_m satisfies (1 + r_m)¹² = 1.07. Taking the 12th root: r_m = 1.07^1/12 − 1. Since 1.07^1/12 ≈ 1.005654, the monthly rate is about 0.005654, or 0.5654%. Check the answer by compounding it back: 1.005654¹² = 1.07, which matches the annual factor exactly.

Notice that 0.5654% is not 7/12 ≈ 0.5833%. Dividing the rate by 12 ignores compounding, which is the whole point of the unit. The same idea lets you rewrite any exponential form: 5^2x = 25^x, and a function given as f(t) = 100(2)^t/12 doubles every 12 time units because the exponent reaches 1 when t = 12.

2.3 Compound Interest

When interest compounds n times per year at annual rate r, the balance after t years is A = P(1 + r/n)^nt. The exponent nt counts the total number of compounding periods, and r/n is the rate per period. The formula requires r as a decimal, so 5% enters as 0.05.

Worked example: $1,000 invested at 5% compounded quarterly for 10 years. Here P = 1000, r = 0.05, n = 4, t = 10, so the per-period rate is 0.05/4 = 0.0125 and the number of periods is 4 × 10 = 40. The balance is A = 1000(1.0125)⁴⁰. Since 1.0125⁴⁰ ≈ 1.643619, the account holds about $1,643.62. To sanity-check the exponent, note that quarterly compounding over 10 years must involve exactly 40 growth steps.

2.4 The Number e and Continuous Growth

The number e ≈ 2.71828 is what the expression (1 + 1/n)ⁿ approaches as n grows without bound. Intuitively, e is the result of compounding 100% growth infinitely often. When compounding is continuous, the compound interest formula A = P(1 + r/n)^nt converges to A = Pe^rt, the continuous growth model.

Worked example: $1,000 at 5% compounded continuously for 10 years. Here A = 1000e^0.05×10 = 1000e^0.5. Since e^0.5 ≈ 1.648721, the balance is about $1,648.72. Compare with the quarterly result of $1,643.62 from the previous section: continuous compounding earns $5.10 more, exactly as you would expect, since compounding more often can only raise the balance for a positive rate.

2.5 Logarithms: Definition and Evaluation

The statement log_b(x) = y means b^y = x, with b > 0, b ≠ 1, and x > 0. The logarithm answers the question "to what power must I raise b to get x?" Two special cases have their own names: log₁₀ is written as log, and log_e is written as ln, the natural logarithm.

To evaluate a logarithm by hand, convert it to the exponential form and solve for the power. Check each answer by confirming b^y = x.

• log₂(32) = 5, because 2⁵ = 32. Confirm: 2 × 2 × 2 × 2 × 2 = 32.
• log₅(1/25) = −2, because 5⁻² = 1/5² = 1/25. Confirm: 1/25 = 0.04, and 5⁻² = 0.04.
• log₃(81) = 4, because 3⁴ = 81. Confirm: 3 × 3 × 3 × 3 = 81.
• log₁₀(0.001) = −3, because 10⁻³ = 1/1000 = 0.001. Confirm: 1/10³ = 1/1000.

Three values are worth memorizing because they fall out of the definition directly: log_b(1) = 0 for any base, since b⁰ = 1; log_b(b) = 1, since b¹ = b; and ln(e) = 1, since e¹ = e.

2.6 Properties of Logarithms

The log properties turn the exponent rules into addition, subtraction, and multiplication. For x > 0 and y > 0: the product rule log_b(xy) = log_b x + log_b y, the quotient rule log_b(x/y) = log_b x − log_b y, and the power rule log_b(x^p) = p · log_b x. The change of base formula log_b x = ln x / ln b lets a calculator evaluate any base.

Worked example, expanding step by step: log₅(25x²/y). First apply the quotient rule to split the division: log₅(25x²) − log₅ y. Then the product rule: log₅ 25 + log₅(x²) − log₅ y. Then the power rule: log₅ 25 + 2 log₅ x − log₅ y. Since log₅ 25 = 2, the expansion is 2 + 2 log₅ x − log₅ y.

Verify it numerically with x = 10 and y = 5. The original expression is log₅(25 × 100 / 5) = log₅(500). By change of base, log₅(500) = ln(500) / ln(5) = 6.214608 / 1.609438 ≈ 3.861353. The expanded form gives 2 + 2 log₅(10) − log₅(5). Since log₅(10) = ln(10)/ln(5) ≈ 1.430677 and log₅(5) = 1, this equals 2 + 2(1.430677) − 1 = 3.861353. Both sides agree.

2.7 Solving Exponential and Logarithmic Equations

To solve an exponential equation, isolate the power and take a logarithm of both sides. For 2^3x−1 = 50, taking log₂ of both sides gives 3x − 1 = log₂(50). Solving for x: x = (1 + log₂ 50) / 3. Evaluate the log with change of base: log₂(50) = ln(50) / ln(2) = 3.912023 / 0.693147 ≈ 5.643856. Then x = (1 + 5.643856) / 3 = 6.643856 / 3 ≈ 2.214619.

Verify by substituting back into the original equation. The exponent becomes 3(2.214619) − 1 = 6.643856 − 1 = 5.643856. Then 2^5.643856 = e^{5.643856 × ln 2} = e^{5.643856 × 0.693147} = e^3.912023 = 50, since e^3.912023 is e^{ln 50}. The solution checks out.

Logarithmic equations need a domain check on every candidate solution, because the log rules can introduce roots that make an argument zero or negative. Solve log x + log(x − 3) = 1 (base 10). Combine with the product rule: log(x(x − 3)) = 1. Convert to exponential form: x(x − 3) = 10¹ = 10, so x² − 3x − 10 = 0, which factors as (x − 5)(x + 2) = 0. The candidates are x = 5 and x = −2.

Check x = 5: log 5 + log 2 = log(5 × 2) = log 10 = 1, so x = 5 works. Check x = −2: the original arguments are log(−2) and log(−5), both undefined, so x = −2 is an extraneous solution and must be rejected. The equation has exactly one solution, x = 5.

2.8 Semi-Log Plots

If data follows an exponential model y = ab^x, taking the log of both sides straightens it into a line. Using log base 10: log y = log(a · b^x) = log a + x log b. This has the form of a line in x with slope log b and y-intercept log a. That is the point of a semi-log plot, which graphs log y against x: exponential data appears as a straight line, and the slope of that line reveals the growth rate.

Read the slope carefully. For y = 4(3)^x, the semi-log equation is log y = log 4 + x log 3, so the slope on a base-10 plot is log 3 ≈ 0.4771, not 3. A positive slope means b > 1 (growth); a negative slope means b < 1 (decay). The intercept gives log a, so a = 10 raised to the intercept value.

2.9 Logistic Models

A logistic model describes growth that is limited by a ceiling, the carrying capacity L: P(t) = L / (1 + Ce^−kt). Early on the model looks exponential, then growth slows as the population approaches L, and the curve flattens out at L. The initial value is P(0) = L / (1 + C), and the growth is fastest at the inflection point, which occurs when P = L/2, half the carrying capacity.

Example: P(t) = 1000 / (1 + 9e^−0.5t). The carrying capacity is 1000, the initial population is P(0) = 1000 / (1 + 9) = 100, and the inflection point is at P = 500. To find when that happens, solve 500 = 1000 / (1 + 9e^−0.5t): dividing gives 1 + 9e^−0.5t = 2, so 9e^−0.5t = 1, e^−0.5t = 1/9, and −0.5t = −ln 9, giving t = 2 ln 9 ≈ 4.394. Check: at t ≈ 4.394, e^−0.5t = e^{−ln 9} = 1/9, so P = 1000 / (1 + 1) = 500, confirming the inflection.

Confusions That Cost Points

Practice Questions

Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next page, so complete the questions before checking them.

Answer Key

1. B. Only B has the variable in the exponent, f(x) = 5(1.3)^x, which is the exponential form a · b^x. A is linear, growing by a constant difference. C is a power function, with the variable as the base, not the exponent. D is a reciprocal function, which decays toward zero rather than multiplying by a constant factor.

2. D. The monthly rate must satisfy (1 + r_m)¹² = 1.06, so r_m = 1.06^1/12 − 1 ≈ 0.004868, or 0.49%. A divides 6% by 12 and ignores compounding. B is 6/100 written as a percent by mistake. C converts to a decimal twice, dividing an already-decimal rate by 12 and calling the result a percent.

3. C. Monthly compounding means n = 12: A = 2000(1 + 0.04/12)⁶⁰ = 2000(1.003333)⁶⁰ ≈ $2,441.99. A uses simple interest, 2000(1 + 0.04 × 5) = $2,400. B compounds annually, 2000(1.04)⁵ ≈ $2,433.31. D has no consistent compounding scheme behind it.

4. A. Continuous growth uses A = Pe^rt with rt = 0.08 × 3 = 0.24, giving 500e^0.24 ≈ 635.62. B compounds once per year instead of continuously. C adds a fixed amount each year, which is linear, not exponential. D puts the growth in the denominator, which models decay rather than growth.

5. C. Set 4^y = 1/16. Since 4² = 16, 4⁻² = 1/16, so y = −2. A and B solve 4^y = 256 and 4^y = 1/256 by misreading the argument. D flips the sign, answering 4² = 16 for a question about 1/16.

6. D. Apply the quotient rule, then the product and power rules: log₃(x²y/z) = log₃(x²y) − log₃ z = 2 log₃ x + log₃ y − log₃ z. A multiplies the logs instead of adding them. B adds log₃ z instead of subtracting it. C divides by log₃ z, as if the quotient rule distributed the division.

7. B. Taking logs gives x = log₃(20) = ln 20 / ln 3 = 2.995732 / 1.098612 ≈ 2.727. A computes ln 20 ≈ 2.996 but forgets to divide by ln 3. C divides the argument by the base, 20/3, which is not a log operation. D takes the reciprocal of the correct answer.

8. A. Combine: log₂(x(x − 2)) = 3, so x(x − 2) = 8, x² − 2x − 8 = 0, and (x − 4)(x + 2) = 0. The candidates are 4 and −2, but x = −2 makes both log arguments negative, so it is extraneous. Only x = 4 survives: log₂ 4 + log₂ 2 = 2 + 1 = 3. B keeps the extraneous root. C fails to check the domain. D makes the argument x − 2 equal to zero, which is undefined.

One-Page Recall Check

Say each answer out loud before you look back, and mark the ones you cannot finish. Anything you cannot say out loud yet belongs in your flashcard deck. In Rycal, add those items to the Exponential and Logarithmic Functions deck and let spaced review bring them back over the next few days.

• Explain the difference between constant percent change and constant difference, with an example of each.
• Convert a 7% annual growth rate to a monthly rate, showing the equation you solve.
• State the compound interest formula and explain what n and the exponent nt count.
• Compute $1,000 at 5% compounded quarterly for 10 years, showing each step.
• Explain where the number e comes from and state the continuous growth formula.
• State the definition of log_b(x) = y in terms of exponents, including the domain restrictions.
• Evaluate log₂(32), log₅(1/25), and log₃(81) by hand, checking each with b^y = x.
• State the product, quotient, and power rules and the change-of-base formula.
• Expand log₅(25x²/y) step by step.
• Solve 2^3x−1 = 50 and verify the solution by substitution.
• Solve log x + log(x − 3) = 1 and explain why one candidate is rejected.
• Explain why a semi-log plot of exponential data is a straight line and what its slope equals.
• Describe a logistic model: carrying capacity, initial value, and where growth is fastest.

Key terms for this unit

Constant percent change, Constant difference, Growth factor, Equivalent exponential forms, Compound interest, Continuous compounding, The number e, Logarithm, Common log, Natural logarithm, Product rule, Quotient rule, Power rule, Change of base, Extraneous solution, Semi-log plot, Logistic model, Carrying capacity, Inflection point.

About this guide. Written for Rycal and aligned to the College Board AP Precalculus course framework, Unit 2. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.