AP Precalculus -2.11 Logarithmic Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.11 Logarithmic Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.11 Logarithmic Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ f $ is given by $ f(x) = \log_3 x $. Which of the following could be the graph of $ y = f^{-1}(x) $?

▶️ Answer/Explanation

Detailed solution

$ f^{-1}(x) = 3^x $, which is an exponential function with base >1.
It passes through (0,1), increases, and has a horizontal asymptote at $ y=0 $ as $ x \to -\infty $.
Only one of the graphs matches this exponential shape.
✅ Answer: (C)

Question

The function $ f $ is an increasing function such that every time the output values of the function $ f $ increase by 1, the corresponding input values multiply by 4. Which of the following could define $ f(x) $?

(A) $ x^4 $
(B) $ 4^x $
(C) $ \sqrt[4]{x} $
(D) $ \log_4 x $

▶️ Answer/Explanation

Detailed solution

If output increases by 1 when input multiplies by 4, then $ f(4x) = f(x) + 1 $.
This is a property of logarithmic functions: $ \log_b(kx) = \log_b k + \log_b x $.
With $ b=4 $, $ \log_4(4x) = \log_4 4 + \log_4 x = 1 + \log_4 x $.
✅ Answer: (D)

Question

The function $ f $ is given by $ f(x) = 2 \log_5 x $. Which of the following describes $ f $?

(A) $ f $ is an increasing function that increases at an increasing rate.
(B) $ f $ is an increasing function that increases at a decreasing rate.
(C) $ f $ is a decreasing function that decreases at an increasing rate.
(D) $ f $ is a decreasing function that decreases at a decreasing rate.

▶️ Answer/Explanation

Detailed solution

$ f(x) = 2 \log_5 x $ is a logarithmic function with base >1, so it is increasing.
Derivative $ f'(x) = \frac{2}{x \ln 5} $ which is positive but decreasing as $ x $ increases, so rate of increase slows down.
Graph is concave down ⇒ increases at a decreasing rate.
✅ Answer: (B)

Question

Which of the following could describe a single logarithmic function $ f $?

(A) $ \lim_{x \to 0^+} f(x) = -\infty $ and $ \lim_{x \to \infty} f(x) = -\infty $
(B) $ \lim_{x \to 0^+} f(x) = -\infty $ and $ \lim_{x \to \infty} f(x) = k $, where $ k $ is a positive constant
(C) $ \lim_{x \to 0^+} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = 0 $
(D) $ \lim_{x \to 0^+} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = -\infty $

▶️ Answer/Explanation

Detailed solution

Typical log function $ \log_b x $ with $ b>1 $ satisfies $ \lim_{x \to 0^+} = -\infty $, $ \lim_{x \to \infty} = \infty $.
But if we take $ f(x) = -\log_b x $ (still logarithmic), then $ \lim_{x \to 0^+} = \infty $, $ \lim_{x \to \infty} = -\infty $.
This matches (D).
✅ Answer: (D)

Question

The logarithmic function $ f $ is defined by $ f(x) = \log_3 x $ on a domain of $ 0 < x \leq 9 $. Which of the following is true of $ f $?

(A) $ f $ has both a maximum and a minimum value.
(B) $ f $ has a maximum value, but no minimum value.
(C) $ f $ has a minimum value, but no maximum value.
(D) $ f $ has neither a minimum value nor a maximum value.

▶️ Answer/Explanation

Detailed solution

$ f $ is increasing on $ (0, 9] $.
As $ x \to 0^+ $, $ f(x) \to -\infty $, so no minimum.
At $ x = 9 $, $ f(9) = \log_3 9 = 2 $, which is the maximum on the closed interval at the right endpoint.
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = \ln x $. Which of the following describes input values for which the output values of $ f $ are integers?

(A) Integer powers of $ e $
(B) Integer powers of 10
(C) Integers raised to the power $ e $
(D) Integers raised to the power 10

▶️ Answer/Explanation

Detailed solution

The natural logarithm $ \ln x $ is the logarithm with base $ e $, so $ \ln(e^k) = k $ for any real number $ k $.
If $ k $ is an integer, then $ \ln(e^k) $ is an integer.
Therefore, input values that are integer powers of $ e $ produce integer outputs.
✅ Answer: (A)

Question

The function $ f $ is logarithmic, and the points $ (2,1) $ and $ (4,2) $ are on the graph of $ f $ in the $ xy $-plane. Which of the following could define $ f(x) $?

(A) $ \log_4 x $
(B) $ 2\log_2 x $
(C) $ 2\log_4 x $
(D) $ \log_4(x + 2) $

▶️ Answer/Explanation

Detailed solution

Check each option against the two given points $ (2,1) $ and $ (4,2) $.
(C) $ f(x) = 2 \log_4 x $:
For $ x = 2 $: $ f(2) = 2 \log_4 2 = 2 \cdot \frac{1}{2} = 1 $ ✅
For $ x = 4 $: $ f(4) = 2 \log_4 4 = 2 \cdot 1 = 2 $ ✅
Both points satisfy the equation.
✅ Answer: (C)

Question

The function $ g $ has the property that for each time the input values double, the output values increase by 1. Which of the following could be the graph of $ y = g(x) $ in the $ xy $-plane?

▶️ Answer/Explanation

Detailed solution

The property describes logarithmic behavior:
If input doubles ($ x \to 2x $), output increases by 1 ($ g(2x) = g(x) + 1 $).
This matches $ g(x) = \log_2 x $ because:
\[ \log_2(2x) = \log_2 2 + \log_2 x = 1 + \log_2 x = 1 + g(x) \]
So $ g $ is logarithmic, with a slowly increasing curve, domain $ x > 0 $, and passes through $(1,0)$, $(2,1)$, $(4,2)$, etc.

✅ Answer: (B)

Question

Which one of the following logarithmic equations has a vertical asymptote of $\lim_{x \to 2^-} f(x) = \infty$?

a. $y = -\log(-x – 2) + 3$

b. $y = -2 \log(-2x + 4) – 2$

c. $y = -\log(x – 2)$

d. $y = \log(-(x – 2)) + 1$

▶️ Answer/Explanation

Detailed solution

The correct option is d.
The limit $\lim_{x \to 2^-} f(x) = \infty$ requires the function to be defined for $x < 2$.
In option (d), the argument is $-(x – 2)$, which is positive when $x < 2$.
As $x$ approaches $2$ from the left ($2^-$), the argument $-(x – 2)$ approaches $0^+$.
The basic property of logarithms is that $\lim_{u \to 0^+} \log(u) = -\infty$.
However, we need the limit to be $+\infty$, which usually requires a reflection.

Question

Consider the functions $ f $ and $ g $ given by $ f(x) = \ln(-x + 8) $ and $ g(x) = \ln(x + 2) + \ln(x – 8) $. In the $ xy $-plane, what are all the $ x $-coordinates of the points of intersection of the graphs $ f $ and $ g $?

(A) $ x = 8 $

(B) $ x = -3 $

(D) No Solution

▶️ Answer/Explanation

Detailed solution

To find the intersection, we must ensure $ x $ exists in the domains of both functions.
First, for $ f(x) = \ln(-x + 8) $, the argument must be positive: $ -x + 8 > 0 \Rightarrow x < 8 $.
Second, for $ g(x) = \ln(x + 2) + \ln(x – 8) $, both arguments must be positive.
This requires $ x > -2 $ and $ x > 8 $. The combined condition for $ g(x) $ is $ x > 8 $.
Comparing the domains, $ f(x) $ requires $ x < 8 $ while $ g(x) $ requires $ x > 8 $.
Since no real number is simultaneously less than 8 and greater than 8, the domains do not overlap.
Therefore, there are no points of intersection. (Note: $ x=-3 $ is an extraneous algebraic solution).

Question

The function $ g $ has the property that for each time the input values double, the output values increase by 1. Which of the following could be the graph of $ g(x) $ in the $ xy $-plane?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The problem describes a function property: when input $ x $ doubles ($ x \rightarrow 2x $), the output increases by 1 ($ g(x) \rightarrow g(x) + 1 $).
This relationship is written as $ g(2x) = g(x) + 1 $, which is characteristic of a logarithmic function $ g(x) = \log_2(x) $.
Let’s test the points on Graph (B) to verify this:
• At $ x = 1 $, $ y = 0 $.
• At $ x = 2 $ (doubled), $ y = 1 $. The increase is $ 1 – 0 = 1 $.
• At $ x = 4 $ (doubled), $ y = 2 $. The increase is $ 2 – 1 = 1 $.
• At $ x = 8 $ (doubled), $ y = 3 $. The increase is $ 3 – 2 = 1 $.
Graphs (C) and (D) represent linear functions passing through the origin, where doubling input doubles the output.
Graph (A) represents an exponential function, which grows much faster than the required condition.
Therefore, Graph (B) is the correct answer.

Question

A logarithmic function $ g $ is created by transforming the function $ f(x) = \log_5(x) $ such that $ g(x) = \frac{1}{2}f(x+3) + 2 $. What is the value of $ g^{-1}(3) $?

(A) $ 55 $

(B) $ \sqrt{7} $

(D) $ 22 $

▶️ Answer/Explanation

Detailed solution

To find the value of $ g^{-1}(3) $, we need to find the value of $ x $ such that $ g(x) = 3 $.
First, substitute $ f(x) $ into the given equation for $ g(x) $:
$ g(x) = \frac{1}{2}\log_5(x+3) + 2 $
Set the equation equal to 3 and solve for $ x $:
$ \frac{1}{2}\log_5(x+3) + 2 = 3 $
Subtract 2 from both sides: $ \frac{1}{2}\log_5(x+3) = 1 $
Multiply both sides by 2: $ \log_5(x+3) = 2 $
Convert the logarithmic equation to exponential form: $ x + 3 = 5^2 $
Solve for $ x $: $ x = 25 – 3 = 22 $.
Therefore, the correct option is (D).

Question

A logarithmic function $ f(x) = 2 \log(x + h) + k $ has an end behavior of $ \lim_{x\to-3^+} f(x) = -\infty $ and passes through the point $ (-2, 4) $. Which of the following are the values for $ h $ and $ k $?

(A) $ h = 3 $ and $ k = 4 $

(B) $ h = 4 $ and $ k = 7 $

(D) $ h = 7 $ and $ k = 8 $

▶️ Answer/Explanation

Detailed solution

The correct option is (A).

The limit $ \lim_{x\to-3^+} f(x) = -\infty $ implies a vertical asymptote at $ x = -3 $.
For the logarithm to be undefined at this asymptote, the argument must be zero: $ x + h = 0 $.
Substituting $ x = -3 $, we get $ -3 + h = 0 $, so $ h = 3 $.
The function passes through $ (-2, 4) $, so substitute $ x = -2 $, $ f(x) = 4 $, and $ h = 3 $.
$ 4 = 2 \log(-2 + 3) + k \Rightarrow 4 = 2 \log(1) + k $.
Since $ \log(1) = 0 $, the equation becomes $ 4 = 0 + k $, which gives $ k = 4 $.

Question

Consider the constant function $ f $ given by $ f(x) = -1 $ and the function $ g $ given by $ g(x) = \log_3 x $. Let $ h $ be the function given by $ h(x) = g(x) – f(x) $. In the $ xy $-plane, what is the $ x $-intercept of the graph of $ h $?

(A) $ (-3, 0) $

(B) $ (\frac{1}{3}, 0) $

(D) The graph of $ h $ does not have an $ x $-intercept.

▶️ Answer/Explanation

Detailed solution

The correct option is (B).

First, determine the expression for $ h(x) $ by substituting $ f(x) $ and $ g(x) $.
Given $ h(x) = g(x) – f(x) $, we have $ h(x) = \log_3 x – (-1) $, which simplifies to $ h(x) = \log_3 x + 1 $.
To find the $ x $-intercept, we set the function equal to zero: $ h(x) = 0 $.
This gives the equation: $ \log_3 x + 1 = 0 $.
Isolate the logarithmic term by subtracting 1 from both sides: $ \log_3 x = -1 $.
Convert the logarithmic equation to exponential form ($ x = b^y $): $ x = 3^{-1} $.
Simplifying this yields $ x = \frac{1}{3} $.
Therefore, the $ x $-intercept is the coordinate $ (\frac{1}{3}, 0) $.

Question (Calc allowed)

The function $S$ is given by $S(t) = \frac{500,000}{1 + 0.4e^{kt}}$, where $k$ is a constant. If $S(4) = 300,000$, what is the value of $S(12)$?

(A) $175,325$
(B) $214,772$
(C) $343,764$
(D) $357,143$

▶️ Answer/Explanation

Detailed solution

Set up the equation $S(4) = 300,000$: $\frac{500,000}{1 + 0.4e^{4k}} = 300,000$.
Simplify to find $e^{4k}$: $1 + 0.4e^{4k} = \frac{500,000}{300,000} = \frac{5}{3}$.
Solve for the exponential term: $0.4e^{4k} = \frac{5}{3} – 1 = \frac{2}{3}$, so $e^{4k} = \frac{2/3}{0.4} = \frac{5}{3}$.
Express the target value $S(12)$ using $e^{4k}$: $S(12) = \frac{500,000}{1 + 0.4(e^{4k})^3}$.
Substitute $e^{4k} = \frac{5}{3}$ into the expression: $S(12) = \frac{500,000}{1 + 0.4(\frac{5}{3})^3}$.
Calculate the denominator: $1 + 0.4(\frac{125}{27}) = 1 + \frac{50}{27} = \frac{77}{27}$.
Final calculation: $S(12) = \frac{500,000 \times 27}{77} \approx 175,324.67$.
The correct option is (A).

Question

The function $g$ has the property that for each time the input values double, the output values increase by $1$. Which of the following could be the graph of $y = g(x)$ in the $xy$-plane?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The given property $g(2x) = g(x) + 1$ defines a logarithmic function base $2$.
In graph (B), when the input $x$ is $1$, the output $y$ is $0$.
Doubling the input to $x = 2$ results in an output of $y = 1$ (an increase of $1$).
Doubling the input again to $x = 4$ results in an output of $y = 2$ (an increase of $1$).
Doubling the input once more to $x = 8$ results in an output of $y = 3$ (an increase of $1$).
This constant arithmetic increase for a geometric change in input is unique to logarithmic curves.
Graphs (C) and (D) are linear, where output increases by a constant amount for a constant addition to the input.
Graph (A) is exponential, which is the inverse relationship where input increases by $1$ as output doubles.
The correct choice is (B).

Question

The function $f$ is an increasing function such that every time the output values of the function $f$ increase by $1$, the corresponding input values multiply by $4$. Which of the following could define $f(x)$?

(A) $x^4$

(B) $4^x$

(D) $\log_{4} x$

▶️ Answer/Explanation

Detailed solution

The problem states that if $f(x) = y$, then $f(4x) = y + 1$.
This relationship is a characteristic property of logarithmic functions.
For $f(x) = \log_{4} x$, if the input is multiplied by $4$, we get $f(4x) = \log_{4}(4x)$.
Using log properties: $\log_{4}(4x) = \log_{4} 4 + \log_{4} x$.
Since $\log_{4} 4 = 1$, the expression becomes $1 + f(x)$.
This matches the condition that the output increases by $1$ when the input is multiplied by $4$.
Therefore, the correct option is (D).

Question

The function $f$ is given by $f(x) = 2 \log_{5} x$. Which of the following describes $f$?

(A) $f$ is an increasing function that increases at an increasing rate.

(B) $f$ is an increasing function that increases at a decreasing rate.

(D) $f$ is a decreasing function that decreases at a decreasing rate.

▶️ Answer/Explanation

Detailed solution

The function is $f(x) = 2 \log_{5} x$, where the base $5 > 1$, making $f$ an increasing function.
To find the rate of change, we calculate the first derivative: $f'(x) = \frac{2}{x \ln 5}$.
Since $f'(x) > 0$ for all $x > 0$, the function is strictly increasing.
To find the rate of that increase, we look at the second derivative: $f”(x) = -\frac{2}{x^2 \ln 5}$.
Since $f”(x) < 0$ for all $x > 0$, the graph is concave down, meaning the rate is decreasing.
Therefore, $f$ increases at a decreasing rate.
The correct option is (B).

Question

Which of the following could describe a single logarithmic function $f$?

(A) $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = -\infty$

(B) $\lim_{x \to 0^+} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = k$, where $k$ is a positive constant

(D) $\lim_{x \to 0^+} f(x) = \infty$ and $\lim_{x \to \infty} f(x) = -\infty$

▶️ Answer/Explanation

Detailed solution

A standard logarithmic function is $f(x) = \log_b(x)$.
If $0 < b < 1$, the function represents a logarithmic decay.
As $x$ approaches $0$ from the right, $\lim_{x \to 0^+} \log_b(x) = \infty$.
As $x$ increases toward infinity, $\lim_{x \to \infty} \log_b(x) = -\infty$.
This end behavior matches the conditions provided in option (D).
Other options fail because logs do not have horizontal asymptotes or return to $-\infty$.
Therefore, the correct description is given by choice (D).

Question

The logarithmic function $f$ is defined by $f(x) = \log_{3} x$ on a domain of $f$ is $0 < x \le 9$. Which of the following is true of $f$ ?

(A) $f$ has both a maximum and a minimum value.

(B) $f$ has a maximum value, but no minimum value.

(D) $f$ has neither a minimum value nor a maximum value.

▶️ Answer/Explanation

Detailed solution

The function $f(x) = \log_{3} x$ is strictly increasing over its entire domain.
At the right endpoint $x = 9$, the function reaches a maximum value of $f(9) = \log_{3} 9 = 2$.
As $x$ approaches the left boundary $0$ from the right, the function $f(x) \to -\infty$.
Because the interval is open at $x = 0$, the function decreases without bound.
Therefore, there is no minimum value because the function never reaches a lowest point.
The correct option is (B).

Question

The function $f$ is given by $f(x) = \ln x$. Which of the following describes input values for which the output values of $f$ are integers?

(A) Integer powers of $e$

(B) Integer powers of $10$

(D) Integers raised to the power $10$

▶️ Answer/Explanation

Detailed solution

The function is defined as $f(x) = \ln x$, where the base is $e$.
To find when the output is an integer, set $f(x) = n$, where $n$ is any integer.
This gives the equation $\ln x = n$.
By converting the logarithmic form to exponential form, we get $x = e^n$.
Therefore, the input values $x$ must be integer powers of $e$.
For example, if $x = e^2$, then $f(e^2) = \ln(e^2) = 2$, which is an integer.
This confirms that option (A) is the correct description.

Question

The function $f$ is logarithmic, and the points $(2, 1)$ and $(4, 2)$ are on the graph of $f$ in the $xy$-plane. Which of the following could define $f(x)$?

(A) $\log_{4} x$
(B) $2 \log_{2} x$
(C) $2 \log_{4} x$
(D) $\log_{4} (x + 2)$

▶️ Answer/Explanation

Detailed solution

To find the correct function, we test the given points $(2, 1)$ and $(4, 2)$ in each option.
For (A): $f(2) = \log_{4} 2 = 0.5$, which does not equal $1$.
For (B): $f(2) = 2 \log_{2} 2 = 2(1) = 2$, which does not equal $1$.
For (C): $f(2) = 2 \log_{4} 2 = 2(0.5) = 1$, which matches the first point.
Checking (C) with the second point: $f(4) = 2 \log_{4} 4 = 2(1) = 2$, which matches $(4, 2)$.
For (D): $f(2) = \log_{4} (2 + 2) = \log_{4} 4 = 1$, but $f(4) = \log_{4} 6 \neq 2$.
Therefore, the correct function is (C) $f(x) = 2 \log_{4} x$.

Question

The function $g$ is given by $g(x) = \log_{3} x$. What is the value of $g\left(\frac{1}{9}\right)$?

(A) $-2$

(B) $-\frac{1}{2}$

(D) $2$

▶️ Answer/Explanation

Detailed solution

To find the value of $g\left(\frac{1}{9}\right)$, substitute $x = \frac{1}{9}$ into the function $g(x) = \log_{3} x$.
This gives the expression $g\left(\frac{1}{9}\right) = \log_{3} \left(\frac{1}{9}\right)$.
Rewrite the fraction as a power of $3$ using the rule $\frac{1}{a^n} = a^{-n}$.
Since $9 = 3^2$, it follows that $\frac{1}{9} = 3^{-2}$.
Substitute this back to get $\log_{3} (3^{-2})$.
Using the property $\log_{b} (b^y) = y$, the expression simplifies to $-2$.
Therefore, the correct value is $-2$, which corresponds to option (A).

Question

Let $h$ be a function such that the output values of $h$ increase without bound as the input values increase without bound, and the output values of $h$ decrease without bound as the input values decrease arbitrarily close to $0$. Which of the following could be the graph of $h$?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The correct graph is (B).
As $x \to \infty$, the output $h(x) \to \infty$, which is shown by the graph rising on the right.
As $x \to 0^+$, the output $h(x) \to -\infty$, indicating a vertical asymptote at $x = 0$.
Graph (A) represents an exponential growth function where $h(x) \to 0$ as $x \to -\infty$.
Graph (C) shows $h(x)$ decreasing as $x$ increases, contradicting the first condition.
Graph (D) shows $h(x) \to -\infty$ as $x \to \infty$, which also contradicts the prompt.
Only graph (B), typical of a logarithmic function $\log_b(x)$ where $b > 1$, fits both criteria.

AP Precalculus -2.11 Logarithmic Functions- MCQ Exam Style Questions - Effective Fall 2023