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AP Precalculus -2.10 Inverses of Exponential Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.10 Inverses of Exponential Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.10 Inverses of Exponential Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

In a certain town, the population in the year 2000 was about \( 30{,}000 \). The population grows at a rate of \( 2.3\% \) per year, and time is measured in years since 2000. Which of the following functions gives output values, in years since 2000, for input values of the town’s population \( p \)?
(A) \( f(p) = 30{,}000 \cdot (1.023)^p \)
(B) \( g(p) = \log_{1.023} \left( \frac{p}{30{,}000} \right) \)
(C) \( h(p) = \frac{p-30{,}000}{2.3} \)
(D) \( k(p) = 30{,}000 \cdot \log_{1.023} p \)
▶️ Answer/Explanation
Detailed solution

Population model: \( P(t) = 30{,}000 \cdot (1.023)^t \) where \( t \) = years since 2000.
We want \( t \) as a function of \( p \): solve \( p = 30{,}000 \cdot (1.023)^t \) for \( t \).
Divide: \( \frac{p}{30{,}000} = (1.023)^t \)
Take \(\log_{1.023}\): \( t = \log_{1.023} \left( \frac{p}{30{,}000} \right) \)
This matches \( g(p) \).
✅ Answer: (B)

Question 

The function \( g \) is given by \( g(x) = \frac{4x+6}{5} \). Which of the following defines \( g^{-1}(x) \)?
(A) \( \frac{5}{4x+6} \)
(B) \( \frac{5x+6}{4} \)
(C) \( \frac{5x}{4} – 6 \)
(D) \( \frac{5x-6}{4} \)
▶️ Answer/Explanation
Detailed solution

Solve \( y = \frac{4x+6}{5} \) for \( x \):
Multiply by 5: \( 5y = 4x + 6 \)
Subtract 6: \( 5y – 6 = 4x \)
Divide by 4: \( x = \frac{5y-6}{4} \)
Thus \( g^{-1}(x) = \frac{5x-6}{4} \).
✅ Answer: (D)

Question 

The exponential function \( g \) is given by \( g(x) = 5^x \). Which of the following expressions defines \( g^{-1}(x) \)?
(A) \( \log_5 x \)
(B) \( \log_x 5 \)
(C) \( \sqrt[5]{x} \)
(D) \( \sqrt[5]{5} \)
▶️ Answer/Explanation
Detailed solution

The inverse of \( y = 5^x \) is \( x = 5^y \) ⇒ \( y = \log_5 x \).
✅ Answer: (A)

Question 

 
 
 
 
 
 
 
 
 
 
 
 
The graph of the exponential function \( f \) is given. Which of the following could be a table of values for the inverse function of \( f \)?
(A) \[\begin{array}{c|c} x & f^{-1}(x) \\ \hline 5 & 25 \\ 2 & 4 \\ \end{array}\]
(B) \[\begin{array}{c|c} x & f^{-1}(x) \\ \hline 25 & 5 \\ 4 & 2 \\ \end{array}\]
(C) \[\begin{array}{c|c} x & f^{-1}(x) \\ \hline 32 & 5 \\ 4 & 2 \\ \end{array}\]
(D) \[\begin{array}{c|c} x & f^{-1}(x) \\ \hline 5 & 32 \\ 2 & 4 \\ \end{array}\]
▶️ Answer/Explanation
Detailed solution

From the graph, \( f \) appears to be \( f(x) = 2^x \).
Then \( f^{-1}(x) = \log_2 x \).
Check values: \( f^{-1}(32) = \log_2 32 = 5 \), \( f^{-1}(4) = \log_2 4 = 2 \).
So table should be \( (32, 5) \) and \( (4, 2) \), which matches option (C).
✅ Answer: (C)

Question 

Which of the following is the inverse of the function \( f \) given by \( f(x) = 4 \log_2(x + 3) – 1 \)?
(A) \( g(x) = \frac{1}{4}(2^x + 1) – 3 \)
(B) \( g(x) = \frac{1}{4} \cdot 2^{(x+1)} – 3 \)
(C) \( g(x) = 2^{(\frac{x}{4}+1)} – 3 \)
(D) \( g(x) = 2^{(\frac{x+1}{4})} – 3 \)
▶️ Answer/Explanation
Detailed solution

Let \( y = 4 \log_2(x+3) – 1 \).
\( y+1 = 4 \log_2(x+3) \)
\( \frac{y+1}{4} = \log_2(x+3) \)
\( x+3 = 2^{(y+1)/4} \)
\( x = 2^{(y+1)/4} – 3 \)
Thus \( f^{-1}(x) = 2^{(x+1)/4} – 3 \).
✅ Answer: (D)

Question 

The initial population size of an animal species is measured to be \( 2000 \). The population doubles every \( 8 \) years. Which of the following functions gives the time, in years, as an output value, and a certain number \( x \) for the population size as an input value?
(A) \( f(x) = \frac{1}{8} \log_2\left(\frac{x}{2000}\right) \)
(B) \( g(x) = \log_2\left(\frac{8x}{2000}\right) \)
(C) \( h(x) = 8 \log_2\left(\frac{x}{2000}\right) \)
(D) \( k(x) = 2000 \log_8 x \)
▶️ Answer/Explanation
Detailed solution

The exponential growth model is: \( P(t) = 2000 \cdot 2^{(t/8)} \), where \( t \) is time in years.
We want the inverse function \( t \) in terms of population \( x \).
Set \( x = 2000 \cdot 2^{(t/8)} \).
Divide both sides by \( 2000 \): \( \frac{x}{2000} = 2^{(t/8)} \).
Take log base 2: \( \log_2\left(\frac{x}{2000}\right) = \frac{t}{8} \).
Multiply by 8: \( t = 8 \log_2\left(\frac{x}{2000}\right) \).
This matches option (C).
✅ Answer: (C)

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