AP Precalculus Logarithmic Functions — Questions and Answers
Question 1: Let f(x) = 2x + 3 and g(x) = x². What is the value of (f ∘ g)(2)?
Correct answer: 11
(f ∘ g)(2) = f(g(2)) = f(4) = 2(4) + 3 = 11. The inner function g is applied first.
In a composition (f ∘ g)(x) = f(g(x)), the function nearest the input is applied first. Here g(2) = 2² = 4, and then f(4) = 2·4 + 3 = 11. Reversing the order gives (g ∘ f)(2) = g(7) = 49, which shows that composition is generally not commutative. Always track which function acts first.
Question 2: Let f(x) = 3·2ˣ. Which expression gives the inverse function f⁻¹(x)?
- log₂(x/3) (Correct answer)
- 3·log₂(x)
- log₂(x) − 3
- log₂(3x)
Correct answer: log₂(x/3)
Set y = 3·2ˣ, divide by 3 to get y/3 = 2ˣ, then take log₂ of both sides: x = log₂(y/3). Swapping variables gives f⁻¹(x) = log₂(x/3).
To invert a function, write y = f(x), swap x and y (or solve for x), and then isolate the new variable. The exponential 3·2ˣ was built by applying 2ˣ and then multiplying by 3. The inverse undoes these steps in reverse order: divide by 3, then apply log₂. Check: f(f⁻¹(x)) = 3·2^(log₂(x/3)) = 3·(x/3) = x.
Question 3: What is the solution to log₃(x + 2) = 2?
Correct answer: 7
Rewriting in exponential form gives x + 2 = 3² = 9, so x = 7.
The statement log_b(A) = C is equivalent to A = b^C. With b = 3 and C = 2, we get x + 2 = 9. Subtracting 2 gives x = 7. Check: log₃(7 + 2) = log₃ 9 = 2, and the argument 9 is positive, so the solution is valid.
Question 4: Which expression is equivalent to log(x³√y / z) for positive x, y, z?
- 3 log x + ½ log y + log z
- 3 log x + ½ log y − log z (Correct answer)
- 3 log x · ½ log y ÷ log z
- (3 log x + ½ log y) ÷ log z
Correct answer: 3 log x + ½ log y − log z
The quotient rule turns division into subtraction, the product rule turns multiplication into addition, and the power rule brings down exponents (√y = y^(1/2)).
Apply log(A/B) = log A − log B to get log(x³√y) − log z. Then log(AB) = log A + log B gives log x³ + log √y − log z. Finally, the power rule log(Aᵖ) = p·log A gives 3 log x + ½ log y − log z. Logarithms never turn products or quotients into products or quotients of logs.
Question 5: What is the domain of f(x) = ln(5 − 2x)?
- x > 5/2
- x < −5/2
- x < 5/2 (Correct answer)
- x ≤ 5/2
Correct answer: x < 5/2
The argument of a logarithm must be strictly positive, so 5 − 2x > 0, which gives x < 5/2.
Solve 5 − 2x > 0. Subtracting 5 gives −2x > −5, and dividing by −2 reverses the inequality, so x < 5/2. At x = 5/2 the argument is 0, and ln 0 is undefined, so the endpoint is excluded. The vertical asymptote of the graph is at x = 5/2.
Question 6: What is the exact value of log₄ 32?
Correct answer: 5/2
Write both as powers of 2: 32 = 2⁵ and 4 = 2². Then log₄ 32 = 5/2 by the change of base formula, or because 4^(5/2) = 2⁵ = 32.
Using change of base, log₄ 32 = log₂ 32 / log₂ 4 = 5/2. Equivalently, we want the exponent c with 4^c = 32. Since 4^c = 2^(2c) and 32 = 2⁵, we need 2c = 5, so c = 5/2. The answer lies between 2 and 3 because 4² = 16 and 4³ = 64.
Question 7: On a semi-log plot, log₁₀(y) is graphed against x, and the points form a straight line passing through (0, 1) and (2, 5). Which exponential model fits the data?
- y = 2·10ˣ
- y = 10·2ˣ
- y = 100·10ˣ
- y = 10·100ˣ (Correct answer)
Correct answer: y = 10·100ˣ
The line has slope (5 − 1)/(2 − 0) = 2 and intercept 1, so log y = 2x + 1. Then y = 10^(2x+1) = 10·(10²)ˣ = 10·100ˣ.
A linear relationship between log y and x signals exponential growth in y. If log y = mx + b, then y = 10^b · (10^m)ˣ, so the intercept gives the initial value 10^b = 10 and the slope gives the base 10^m = 100. Check at x = 2: 10·100² = 100,000, and log(100,000) = 5, matching the point (2, 5).
Question 8: What is the vertical asymptote of g(x) = log₂(x + 4) − 5?
- x = 4
- x = −4 (Correct answer)
- x = −5
- x = 5
Correct answer: x = −4
The graph of log₂ x has its asymptote at x = 0. Replacing x with x + 4 shifts it left 4 units, giving x = −4. The −5 only moves the graph vertically.
A logarithmic function's vertical asymptote occurs where its argument equals 0. Setting x + 4 = 0 gives x = −4. Horizontal shifts move the asymptote, while vertical shifts such as −5 do not. The domain is x > −4.
Question 9: What is the solution to e^(2x) = 7?
- ln 7 − 2
- ln(7/2)
- (ln 7)/2 (Correct answer)
- ln 14
Correct answer: (ln 7)/2
Take the natural log of both sides: 2x = ln 7, so x = (ln 7)/2.
Because ln and e^x are inverses, ln(e^(2x)) = 2x. Applying ln to both sides gives 2x = ln 7. Dividing by 2 gives x = (ln 7)/2 ≈ 0.973. The 2 is a multiplier on the exponent, so it must be divided out, not subtracted inside the logarithm.
Question 10: Which function is the inverse of h(x) = eˣ + 1?
- h⁻¹(x) = ln(x − 1) (Correct answer)
- h⁻¹(x) = ln(x) − 1
- h⁻¹(x) = ln(x) + 1
- h⁻¹(x) = 1/(eˣ + 1)
Correct answer: h⁻¹(x) = ln(x − 1)
Set y = eˣ + 1, subtract 1 to get y − 1 = eˣ, then take ln: x = ln(y − 1). Swapping variables gives h⁻¹(x) = ln(x − 1).
The function h first applies eˣ and then adds 1, so the inverse must subtract 1 first and then apply ln. Check: h(h⁻¹(x)) = e^(ln(x−1)) + 1 = (x − 1) + 1 = x. The inverse's domain is x > 1, which matches the range of h. Note that the reciprocal 1/h(x) is not the inverse function.
