AP PreCalculus Practice Quiz: Exponential and Logarithmic Equations and Inequalities
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 15 questions to check your progress.
Question 1 of 15
All Questions (15)
A) x = 1
B) x = 2
C) x = 3
D) x = 5
Correct Answer: B
To solve the equation, rewrite both sides with a common base. Since 27 = 3^3, the equation becomes 3^(2x-1) = 3^3. By the property of exponents, if the bases are equal, the exponents must be equal. Therefore, 2x - 1 = 3. Adding 1 to both sides gives 2x = 4, and dividing by 2 gives x = 2.
A) f⁻¹(x) = log₅((x-3)/2) + 1
B) f⁻¹(x) = log₅((x-2)/3) - 1
C) f⁻¹(x) = log₂((x-3)/5) + 1
D) f⁻¹(x) = log₅(x-3) - 2
Correct Answer: A
To find the inverse of y = 2 * 5^(x-1) + 3, first swap x and y: x = 2 * 5^(y-1) + 3. Then, solve for y by reversing the operations. Subtract 3: x - 3 = 2 * 5^(y-1). Divide by 2: (x - 3)/2 = 5^(y-1). Take the logarithm base 5 of both sides: log₅((x - 3)/2) = y - 1. Finally, add 1: y = log₅((x - 3)/2) + 1.
A) x = 25
B) x = -4
C) x = 25 and x = -4
D) No solution
Correct Answer: A
First, use the product property of logarithms to condense the left side: log(x(x - 21)) = 2. Convert the logarithmic equation to its exponential form: x(x - 21) = 10^2. This simplifies to x² - 21x = 100, or x² - 21x - 100 = 0. Factoring the quadratic gives (x - 25)(x + 4) = 0. The potential solutions are x = 25 and x = -4. However, we must check for extraneous solutions. The domain of the original equation requires x > 0 and x - 21 > 0 (i.e., x > 21). The solution x = -4 is extraneous because it is not in the domain. Therefore, the only valid solution is x = 25.
A) g⁻¹(x) = 2^((x+5)/4) - 3
B) g⁻¹(x) = 4^((x-5)/2) + 3
C) g⁻¹(x) = 2^(x+5) - 12
D) g⁻¹(x) = (1/4) * 2^(x+5) + 3
Correct Answer: A
To find the inverse of y = 4log₂(x + 3) - 5, swap x and y: x = 4log₂(y + 3) - 5. Solve for y by reversing the operations. Add 5: x + 5 = 4log₂(y + 3). Divide by 4: (x + 5)/4 = log₂(y + 3). Convert to exponential form: 2^((x+5)/4) = y + 3. Finally, subtract 3: y = 2^((x+5)/4) - 3.
A) 2 + 5log₃(x) - 2log₃(y)
B) 18 + 5log₃(x) - 2log₃(y)
C) 2(5log₃(x) - log₃(y))
D) log₃(9) + 5log₃(x) + 2log₃(y)
Correct Answer: A
Using the properties of logarithms, we can expand the expression. The logarithm of a quotient is the difference of the logarithms: log₃(9x⁵) - log₃(y²). The logarithm of a product is the sum of the logarithms: log₃(9) + log₃(x⁵) - log₃(y²). The logarithm of a power is the exponent times the logarithm: log₃(9) + 5log₃(x) - 2log₃(y). Since log₃(9) = 2, the final expression is 2 + 5log₃(x) - 2log₃(y).
A) x < 9
B) x > 1
C) 1 < x < 9
D) 1 < x < 7
Correct Answer: C
To solve a logarithmic inequality, two conditions must be met. First, the argument of the logarithm must be positive: x - 1 > 0, which means x > 1. Second, we solve the inequality itself. Convert log₂(x - 1) < 3 to exponential form. Since the base (2) is greater than 1, the inequality direction is preserved: x - 1 < 2³. This simplifies to x - 1 < 8, or x < 9. The final solution must satisfy both conditions (x > 1 and x < 9), so the solution is the intersection of these two intervals: 1 < x < 9.
A) f⁻¹(x) = xᵇ
B) f⁻¹(x) = logₓ(b)
C) f⁻¹(x) = logₑ(bˣ)
D) f⁻¹(x) = logₑ(x)
Correct Answer: D
By definition, the inverse of an exponential function y = bˣ is the logarithmic function with the same base. To find the inverse, we swap x and y to get x = bʸ and then solve for y. This gives y = logₑ(x). Therefore, the inverse function is f⁻¹(x) = logₑ(x).
A) To ensure the solution is an integer.
B) To account for rounding errors during calculation.
C) To discard extraneous solutions that fall outside the domain of the original logarithmic expressions.
D) To verify that the properties of logarithms were applied correctly.
Correct Answer: C
The process of solving logarithmic equations, particularly when condensing logarithms, can introduce solutions that are not valid in the original equation. This is because the domain of a logarithmic function logₑ(u) requires that the argument u be positive. A potential solution is considered extraneous if it causes the argument of any logarithm in the original equation to be zero or negative. Therefore, all potential solutions must be checked against the domain limitations of the original equation.
A) x = ln(2), x = ln(4)
B) x = 2, x = 4
C) x = ln(8), x = ln(6)
D) x = e², x = e⁴
Correct Answer: A
This equation is quadratic in form. Let u = eˣ. Then the equation can be rewritten as u² - 6u + 8 = 0. Factoring this quadratic equation gives (u - 2)(u - 4) = 0. The solutions for u are u = 2 and u = 4. Now, substitute back eˣ for u. We have two equations: eˣ = 2 and eˣ = 4. To solve for x, take the natural logarithm of both sides of each equation: x = ln(2) and x = ln(4). Both solutions are valid.
A) x > 1
B) x < 1
C) x > 5
D) x < 5
Correct Answer: A
To solve the inequality, first express both sides with the same base. Since 1/25 = 1/5² = 5⁻², the inequality becomes 5^(x-3) > 5⁻². Because the base (5) is greater than 1, we can compare the exponents directly without changing the inequality direction: x - 3 > -2. Adding 3 to both sides gives x > 1.
A) Reflected over the x-axis, shifted left 4 units, and down 1 unit.
B) Reflected over the y-axis, shifted right 4 units, and up 1 unit.
C) Reflected over the x-axis, shifted right 4 units, and down 1 unit.
D) Reflected over the y-axis, shifted left 4 units, and up 1 unit.
Correct Answer: A
Comparing f(x) = -log₂(x+4) - 1 to the general form f(x) = a logₑ(x-h) + k, we can identify the transformations. The negative sign in front (a = -1) causes a reflection over the x-axis. The term (x+4) corresponds to (x-h), so h = -4, which represents a horizontal shift 4 units to the left. The term -1 corresponds to k, which represents a vertical shift 1 unit down.
A) Take the logarithm of both sides.
B) Rewrite both sides of the equation with a common base of 2.
C) Expand the exponents to get 4x + 4 = 8x - 8.
D) Graph both y = 4^(x+1) and y = 8^(x-1) to find their intersection.
Correct Answer: B
While taking the logarithm of both sides is a valid method, the most direct analytical approach is to use the property of exponents by finding a common base. Since 4 = 2² and 8 = 2³, the equation can be rewritten as (2²)^(x+1) = (2³)^(x-1). This simplifies the equation to 2^(2x+2) = 2^(3x-3), allowing the exponents to be set equal to each other.
A) x = 3
B) x = -1
C) x = 3 and x = -1
D) No solution
Correct Answer: A
First, use the power property of logarithms on the left side: ln(x²) = ln(2x + 3). Since the logarithms have the same base, we can set their arguments equal: x² = 2x + 3. Rearrange this into a quadratic equation: x² - 2x - 3 = 0. Factoring gives (x - 3)(x + 1) = 0, with potential solutions x = 3 and x = -1. We must check these solutions in the original equation's domain. The domain requires x > 0 and 2x + 3 > 0. The solution x = -1 is extraneous because ln(-1) is undefined. The solution x = 3 is valid. Therefore, the only solution is x = 3.
A) f⁻¹(x) = log₄(x) - 1 + 5
B) f⁻¹(x) = log₄(x-5) + 1
C) f⁻¹(x) = log₅(x-1) + 4
D) f⁻¹(x) = log₄(x+1) - 5
Correct Answer: B
This question combines understanding transformations and finding the inverse. To find the inverse of y = 4^(x-1) + 5, we reverse the mapping. First, swap x and y: x = 4^(y-1) + 5. Then, isolate y. Subtract 5: x - 5 = 4^(y-1). Apply the inverse operation, which is log base 4: log₄(x - 5) = y - 1. Finally, add 1: y = log₄(x - 5) + 1.
A) 0.14
B) 0.78
C) 1.08
D) 1.38
Correct Answer: C
This question requires the use of properties of logarithms. First, express 12 as a product of 2s and 3s: 12 = 4 * 3 = 2² * 3. Now, apply the logarithm properties: logₐ(12) = logₐ(2² * 3). Using the product rule: logₐ(2²) + logₐ(3). Using the power rule: 2 * logₐ(2) + logₐ(3). Now substitute the given values: 2 * (0.3) + 0.48 = 0.6 + 0.48 = 1.08.