AP Precalculus -2.12 Logarithmic Function Manipulation- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.12 Logarithmic Function Manipulation- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.12 Logarithmic Function Manipulation- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The functions $ f $ and $ g $ are given by $ f(x) = 4^{(5x-1)} $ and $ g(x) = 8^{(x/4)} $. When solving the equation $ f(x) = g(x) $, the functions can be rewritten in equivalent forms so that the equation can be solved without the use of technology. Which of the following are equivalent definitions of $ f $ and $ g $ that aid in solving $ f(x) = g(x) $ without the use of technology?

(A) $ f(x) = 2^{(\log_2 4 \cdot (5x-1))} $ and $ g(x) = 2^{(\log_2 8 \cdot (x/4))} $
(B) $ f(x) = 2^{(\log_2 8 \cdot (5x-1))} $ and $ g(x) = 2^{(\log_2 4 \cdot (x/4))} $
(C) $ f(x) = 4^{(\log_2 4 – (5x-1))} $ and $ g(x) = 8^{(\log_2 8 \cdot (x/4))} $
(D) $ f(x) = 2 \cdot 4^{(\log_2 4 – (5x-1))} $ and $ g(x) = 8^{(\log_2 8 \cdot (x/4))} $

▶️ Answer/Explanation

Detailed solution

Rewrite $ 4 $ and $ 8 $ as powers of $ 2 $:
$ 4 = 2^2 $, $ 8 = 2^3 $.
So $ f(x) = (2^2)^{5x-1} = 2^{2(5x-1)} = 2^{10x-2} $.
Also $ g(x) = (2^3)^{x/4} = 2^{3x/4} $.
Comparing with the options, $ \log_2 4 = 2 $ and $ \log_2 8 = 3 $.
Thus $ f(x) = 2^{(\log_2 4 \cdot (5x-1))} $ and $ g(x) = 2^{(\log_2 8 \cdot (x/4))} $.
This matches option (A).
✅ Answer: (A)

Question

To solve the equation $ \log_8(x – 3) + \log_8(x + 4) = 1 $, one method is to apply the properties of logarithms to write a new equation that can be used to identify possible solutions. Of the following, which is such an equation?

(A) $ 2x + 1 = 8 $
(B) $ \frac{x-3}{x+4} = 8 $
(C) $ x^2 – 12 = 8 $
(D) $ x^2 + x – 12 = 8 $

▶️ Answer/Explanation

Detailed solution

Combine logs: $ \log_8[(x-3)(x+4)] = 1 $.
Convert to exponential: $ (x-3)(x+4) = 8^1 $.
Expand: $ x^2 + x – 12 = 8 $.
✅ Answer: (D)

Question

An equation involves the expression $ \log_9(27^x) $, which is equivalent to a rational multiple of $ x $. By rewriting the expression in an equivalent form, the value of the rational number can be determined without use of a calculator or complicated calculations. Which of the following is an equivalent expression that satisfies this requirement?

(A) $ x \ln\left(\frac{27}{9}\right) $
(B) $ x \log_3\left(\frac{27}{9}\right) $
(C) $ \frac{x \ln 27}{\ln 9} $
(D) $ \frac{x \log_3 27}{\log_3 9} $

▶️ Answer/Explanation

Detailed solution

Change to base 3: $ \log_9(27^x) = \frac{\log_3(27^x)}{\log_3 9} = \frac{x \log_3 27}{\log_3 9} $.
Since $ \log_3 27 = 3 $ and $ \log_3 9 = 2 $, the expression simplifies to $ \frac{3}{2}x $.
✅ Answer: (D)

Question

If $ m = \log_3 81 $, which of the following is also true?

(A) $ 3m = 81 $
(B) $ 3^m = 81 $
(C) $ \sqrt[3]{m} = 81 $
(D) $ \sqrt[3]{81} = m $

▶️ Answer/Explanation

Detailed solution

By definition of logarithm: $ m = \log_3 81 $ means $ 3^m = 81 $.
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = 9 \cdot 25^x $. Which of the following is an equivalent form for $ f(x) $?

(A) $ f(x) = 3 \cdot 5^{(x/2)} $
(B) $ f(x) = 3 \cdot 5^{2x} $
(C) $ f(x) = 9 \cdot 5^{(x/2)} $
(D) $ f(x) = 9 \cdot 5^{2x} $

▶️ Answer/Explanation

Detailed solution

$ 25 = 5^2 $, so $ 25^x = (5^2)^x = 5^{2x} $.
Thus $ f(x) = 9 \cdot 5^{2x} $.
✅ Answer: (D)

Question

The function $ f $ is given by $ f(x) = \log_2 x $. Which of the following is equivalent to $ f(7) $?

(A) $ \log_{10} \left( \frac{7}{2} \right) $
(B) $ \frac{\log_{10} 2}{\log_{10} 7} $
(C) $ \frac{\log_7 10}{\log_2 10} $
(D) $ \frac{\log_3 7}{\log_3 2} $

▶️ Answer/Explanation

Detailed solution

We have $ f(7) = \log_2 7 $. Using the change of base formula for logarithms:
$ \log_b x = \frac{\log_a x}{\log_a b} $, where $ a > 0 $ and $ a \neq 1 $.
Applying this to $ \log_2 7 $, we can choose any valid base $ a $.
For option (D): $ \frac{\log_3 7}{\log_3 2} = \log_2 7 $, which matches $ f(7) $.
✅ Answer: (D)

Question

The function $ g $ is given by $ g(x) = \log_7 x $, and the function $ h $ is given by $ h(x) = \log_{49} x $. Which of the following describes the relationships between $ g $ and $ h $?

(A) For equal input values, the output values of $ h $ are half the output values of $ g $.
(B) For equal input values, the output values of $ h $ are twice the output values of $ g $.
(C) For equal input values, the output values of $ h $ are the square of the output values of $ g $.
(D) For equal input values, the output values of $ h $ are the square root of the output values of $ g $.

▶️ Answer/Explanation

Detailed solution

Using the change of base formula:
$ h(x) = \log_{49} x = \frac{\log_7 x}{\log_7 49} $.
Since $ 49 = 7^2 $, $ \log_7 49 = 2 $.
Thus, $ h(x) = \frac{\log_7 x}{2} = \frac{g(x)}{2} $.
Therefore, for the same input $ x $, the output of $ h $ is half the output of $ g $.
✅ Answer: (A)

Question

The function $ h $ is given by $ h(x) = \log_3 x $. Which of the following is equivalent to the expression $ 2 \cdot h(w) + h(p) $, where $ w $ and $ p $ are values in the domain of $ h $?

(A) $ \log_3 ((wp)^2) $
(B) $ (\log_3 w)^2 \cdot (\log_3 p) $
(C) $ \log_3 (w^2p) $
(D) $ \log_3 (2wp) $

▶️ Answer/Explanation

Detailed solution

Given $ h(x) = \log_3 x $, so $ h(w) = \log_3 w $ and $ h(p) = \log_3 p $.
The expression becomes:
$ 2 \cdot h(w) + h(p) = 2 \log_3 w + \log_3 p $.
Using the power property $ a \log_b m = \log_b (m^a) $ and the product property $ \log_b m + \log_b n = \log_b (mn) $:
$ 2 \log_3 w = \log_3 (w^2) $
$ \log_3 (w^2) + \log_3 p = \log_3 (w^2p) $.
✅ Answer: (C)

Question

The function $ f $ is given by $ f(x) = \log_{10} x $. The function $ g $ is given by $ g(x) = \log_{10} (x^3) $. Which of the following describes a transformation for which the graph of $ g $ is the image of the graph of $ f $?

(A) A vertical dilation by a factor of 3
(B) A vertical dilation by a factor of $\frac{1}{3}$
(C) A horizontal dilation by a factor of 3
(D) A horizontal dilation by a factor of $\frac{1}{3}$

▶️ Answer/Explanation

Detailed solution

Using the power property for logarithms:
$ g(x) = \log_{10} (x^3) = 3 \log_{10} x $.
Since $ f(x) = \log_{10} x $, we have $ g(x) = 3f(x) $.
Multiplying the output of $ f $ by 3 corresponds to a vertical dilation (stretch) by a factor of 3.
✅ Answer: (A)

Question

The function $ g $ is given by $ g(x) = a \log_b c $, where $ a $, $ b $, and $ c $ are positive integers. Which of the following is an equivalent representation of $ g(x) $?

(A) $ \log_b(c^a) $
(B) $ (\log_b c)^a $
(C) $ \log_b(c^{(1/a)}) $
(D) $ a \log_{10} b + a \log_{10} c $

▶️ Answer/Explanation

Detailed solution

Using the power property for logarithms: $ a \log_b c = \log_b (c^a) $.
Therefore, $ g(x) = \log_b (c^a) $.
This matches option (A).
✅ Answer: (A)

Question

Let $ x $ and $ y $ be positive constants. Which of the following is equivalent to $ 2 \ln x – 3 \ln y $?

(A) $ \ln\left(\frac{x^2}{y^3}\right) $
(B) $ \ln(x^2 y^3) $
(C) $ \ln(2x – 3y) $
(D) $ \ln\left(\frac{2x}{3y}\right) $

▶️ Answer/Explanation

Detailed solution

Using logarithm properties:
$ 2 \ln x = \ln(x^2) $
$ 3 \ln y = \ln(y^3) $
Subtraction of logs corresponds to division:
$ 2 \ln x – 3 \ln y = \ln(x^2) – \ln(y^3) = \ln\left(\frac{x^2}{y^3}\right) $.
✅ Answer: (A)

Question

Let $ k, w, $ and $ z $ be positive constants. Which of the following is equivalent to $ \log_{10}\left(\frac{kz}{w^2}\right) $?

(A) $ \log_{10}(k + z) – \log_{10}(2w) $
(B) $ \log_{10}k + \log_{10}z – 2\log_{10}w $
(C) $ \log_{10}k + \log_{10}z – \frac{1}{2}\log_{10}w $
(D) $ \log_{10}k – \log_{10}z + 2\log_{10}w $

▶️ Answer/Explanation

Detailed solution

Apply logarithm properties step by step:
$ \log_{10}\left(\frac{kz}{w^2}\right) = \log_{10}(kz) – \log_{10}(w^2) $ (quotient rule)
$ = \left(\log_{10}k + \log_{10}z\right) – 2\log_{10}w $ (product rule and power rule).
This matches option (B).
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = a \cdot c^x $, where $ a > 0 $ and $ c > 1 $. Which of the following is true about the values of constants $ m $ and $ b $ in the equation $ \ln(f(x)) = mx + b $?

(A) $ m > 0 $ because $ \ln c > 0 $; $ b $ can be any real number because $ \ln a $ can be any real number.
(B) $ m > 0 $ because $ \ln c > 0 $; $ b > 0 $ because $ \ln a > 0 $.
(C) $ m $ can be any real number because $ \ln c $ can be any real number; $ b $ can be any real number because $ \ln a $ can be any real number.
(D) $ m $ can be any real number because $ \ln c $ can be any real number; $ b > 0 $ because $ \ln a > 0 $.

▶️ Answer/Explanation

Detailed solution

Given $ f(x) = a \cdot c^x $, take the natural log of both sides:
$ \ln(f(x)) = \ln(a \cdot c^x) = \ln a + x \ln c $.
This is in the form $ \ln(f(x)) = mx + b $ with $ m = \ln c $ and $ b = \ln a $.
Since $ c > 1 $, $ \ln c > 0 $, so $ m > 0 $.
Since $ a > 0 $, $ \ln a $ can be any real number (positive, negative, or zero).
Thus, $ b $ can be any real number.
✅ Answer: (A)

Question

Which of the following expressions is equivalent to $\log_{3}(x^{5})$?

(A) $\log_{3}5+\log_{3}x$
(B) $\log_{3}5\cdot \log_{3}x$
(C) $5\log_{3}x$
(D) $\frac{\log_{3}x}{\log_{3}5}$

▶️ Answer/Explanation

Detailed solution

1. Apply Logarithm Power Rule:
The power property states $\log_b(m^p) = p \cdot \log_b m$.

2. Rewrite the expression:
$\log_{3}(x^{5}) = 5\log_{3}x$

✅ Answer: (C)

Question

Let $ x $ and $ y $ be positive constants. Which of the following is equivalent to $ 2\ln x – 3\ln y $?

(A) $ \displaystyle \ln\left(\frac{x^2}{y^3}\right) $
(B) $ \displaystyle \ln(x^2y^3) $
(C) $ \displaystyle \ln(2x – 3y) $
(D) $ \displaystyle \ln\left(\frac{2x}{3y}\right) $

▶️ Answer/Explanation

Detailed solution

Using logarithmic properties:
\[ 2\ln x = \ln(x^2) \] \[ 3\ln y = \ln(y^3) \] So, \[ 2\ln x – 3\ln y = \ln(x^2) – \ln(y^3) \] By the quotient property: \[ \ln(x^2) – \ln(y^3) = \ln\left(\frac{x^2}{y^3}\right) \]
✅ Answer: (A)

Question

Condense the expression $4 \log_{7} x + 4 \log_{7} z – 24 \log_{7} y$ into a single logarithm.

a. $\log_{7}(z^{6} \sqrt{yx})$

b. $\log_{7} \frac{x^{6}}{zy^{4}}$

c. $\log_{7} \frac{z^{4}x^{4}}{y^{24}}$

d. $\log_{7}(yz^{6} \sqrt{x})$

▶️ Answer/Explanation

Detailed solution

The correct option is c.
Apply the Power Property: $4 \log_{7} x = \log_{7} x^{4}$, $4 \log_{7} z = \log_{7} z^{4}$, and $24 \log_{7} y = \log_{7} y^{24}$.
Rewrite the expression: $\log_{7} x^{4} + \log_{7} z^{4} – \log_{7} y^{24}$.
Apply the Product Property to the addition: $\log_{7} (x^{4} \cdot z^{4}) – \log_{7} y^{24}$.
Apply the Quotient Property to the subtraction: $\log_{7} \frac{x^{4}z^{4}}{y^{24}}$.
This matches the expression in option c: $\log_{7} \frac{z^{4}x^{4}}{y^{24}}$.

Question

If $\log_{7} 11 = X$, $\log_{7} 8 = Y$, and $\log_{7} 12 = Z$, then rewrite $\log_{7} \frac{56}{121}$ in terms of the variables given using the properties of logarithms.

a. $1 + Y – X^{2}$

b. $2Y + 2X$

c. $1 + Y – 2X$

d. $1 + Y + 2X$

▶️ Answer/Explanation

Detailed solution

First, use the quotient rule: $\log_{7} \frac{56}{121} = \log_{7} 56 – \log_{7} 121$.
Factor the terms: $56 = 7 \times 8$ and $121 = 11^{2}$.
Apply the product rule: $\log_{7}(7 \times 8) = \log_{7} 7 + \log_{7} 8$.
Simplify the constant: $\log_{7} 7 = 1$.
Apply the power rule: $\log_{7} 11^{2} = 2\log_{7} 11$.
Substitute the variables: $1 + Y – 2X$.
The correct option is c.

Question

The logarithmic function is defined by $\log(y – A) = Bx – \log c$, where $A$, $B$, and $C$ are constant positive numbers. What is the value of $y$ in terms of $x$?

(A) $y = \frac{10^{Bx}}{C} + A$

(B) $y = C + A(10^{Bx})$

(D) $y = \frac{Bx^{10}}{C} + A$

▶️ Answer/Explanation

Detailed solution

The given equation is $\log(y – A) = Bx – \log C$.
Rearrange the terms to group the logarithms: $\log(y – A) + \log C = Bx$.
Apply the product rule $\log m + \log n = \log(mn)$: $\log[C(y – A)] = Bx$.
Convert from logarithmic to exponential form (base 10): $C(y – A) = 10^{Bx}$.
Divide both sides by $C$: $y – A = \frac{10^{Bx}}{C}$.
Add $A$ to both sides to solve for $y$: $y = \frac{10^{Bx}}{C} + A$.
Therefore, the correct option is (A).

Question

The function $ f $ is given by $ f(x) = \log_2 x $, and the function $ g $ is given by $ g(x) = \log_2(8x) $. Consider the graphs of both functions in the same $ xy $-plane. Which of the following statements with reasoning is true?

(A) The graph of $ g $ is a vertical dilation of the graph of $ f $ because $ \log_2(8x) = \log_2 8 + \log_2 x $.

(B) The graph of $ g $ is a vertical dilation of the graph of $ f $ because $ \log_2(8x) = \log_2 8 \cdot \log_2 x $.

(C) The graph of $ g $ is a vertical translation of the graph of $ f $ because $ \log_2(8x) = \log_2 8 + \log_2 x $.

(D) The graph of $ g $ is a vertical translation of the graph of $ f $ because $ \log_2(8x) = \log_2 8 \cdot \log_2 x $.

▶️ Answer/Explanation

Detailed solution

The correct option is (C).

First, apply the product property of logarithms: $ \log_b(mn) = \log_b m + \log_b n $.
Using this property on $ g(x) $, we get: $ g(x) = \log_2(8x) = \log_2 8 + \log_2 x $.
Next, evaluate the constant term: since $ 2^3 = 8 $, it follows that $ \log_2 8 = 3 $.
Substituting this value back into the equation gives: $ g(x) = 3 + \log_2 x $, or $ g(x) = 3 + f(x) $.
In function transformations, adding a constant $ k $ (where $ f(x) + k $) results in a vertical translation.
Therefore, the graph of $ g $ is a vertical translation of the graph of $ f $ shifted up by 3 units.

Question 28

28. To solve the equation $\log_{8}(x – 3) + \log_{8}(x + 4) = 1$, one method is to apply the properties of logarithms to write a new equation that can be used to identify possible solutions. Of the following, which is such an equation?

(A) $2x + 1 = 8$
(B) $\frac{x – 3}{x + 4} = 8$
(C) $x^{2} – 12 = 8$
(D) $x^{2} + x – 12 = 8$

▶️ Answer/Explanation

Detailed solution

The correct option is (D).
Use the product property of logarithms: $\log_{b}(m) + \log_{b}(n) = \log_{b}(m \cdot n)$.
Apply this to the given equation: $\log_{8}((x – 3)(x + 4)) = 1$.
Rewrite the logarithmic equation in exponential form: $(x – 3)(x + 4) = 8^{1}$.
Expand the left side using the FOIL method: $x^{2} + 4x – 3x – 12 = 8$.
Simplify the linear terms to get the final equation: $x^{2} + x – 12 = 8$.

Question

The function $f$ is given by $f(x) = \log_{2} x$. Which of the following is equivalent to $f(7)$?

(A) $\log_{10} \left( \frac{7}{2} \right)$

(B) $\frac{\log_{10} 2}{\log_{10} 7}$

(D) $\frac{\log_{3} 7}{\log_{3} 2}$

▶️ Answer/Explanation

Detailed solution

The function is defined as $f(x) = \log_{2} x$, so we need to find an expression equivalent to $f(7) = \log_{2} 7$.
Using the Change of Base Formula: $\log_{b} a = \frac{\log_{c} a}{\log_{c} b}$ for any positive base $c$.
By applying this formula with a new base $c = 3$, we get $\log_{2} 7 = \frac{\log_{3} 7}{\log_{3} 2}$.
Option (A) is incorrect because $\log_{10} \left( \frac{7}{2} \right) = \log_{10} 7 – \log_{10} 2$.
Option (B) is the reciprocal of the correct change of base result using base $10$.
Option (C) is incorrect because $\frac{\log_{7} 10}{\log_{2} 10} = \frac{1/\log_{10} 7}{1/\log_{10} 2} = \frac{\log_{10} 2}{\log_{10} 7} = \log_{7} 2$.
Therefore, the correct equivalent expression is provided in option (D).
The final answer is (D).

Question

The function $g$ is given by $g(x) = \log_{7} x$, and the function $h$ is given by $h(x) = \log_{49} x$. Which of the following describes the relationships between $g$ and $h$?

(A) For equal input values, the output values of $h$ are half the output values of $g$.

(B) For equal input values, the output values of $h$ are twice the output values of $g$.

(D) For equal input values, the output values of $h$ are the square root of the output values of $g$.

▶️ Answer/Explanation

Detailed solution

The correct answer is (A).
Start with the function $h(x) = \log_{49} x$.
Recognize that the base $49$ can be written as $7^{2}$.
Apply the change of base formula: $\log_{a^n} x = \frac{1}{n} \log_{a} x$.
This gives $h(x) = \log_{7^{2}} x = \frac{1}{2} \log_{7} x$.
Substitute $g(x)$ into the equation to get $h(x) = \frac{1}{2} g(x)$.
Therefore, for the same $x$, the output of $h$ is half the output of $g$.

Question

The function $h$ is given by $h(x) = \log_{3} x$. Which of the following is equivalent to the expression $2 \cdot h(w) + h(p)$, where $w$ and $p$ are values in the domain of $h$?

(A) $\log_{3}((wp)^{2})$
(B) $(\log_{3} w)^{2} \cdot (\log_{3} p)$
(C) $\log_{3}(w^{2}p)$
(D) $\log_{3}(2wp)$

▶️ Answer/Explanation

Detailed solution

Given the function $h(x) = \log_{3} x$, substitute $w$ and $p$ into the expression: $2 \log_{3} w + \log_{3} p$.
Apply the Power Property of logarithms, $a \log_{b} c = \log_{b}(c^{a})$, to rewrite the first term as $\log_{3}(w^{2})$.
The expression becomes $\log_{3}(w^{2}) + \log_{3} p$.
Apply the Product Property of logarithms, $\log_{b} m + \log_{b} n = \log_{b}(mn)$.
Combining the terms results in $\log_{3}(w^{2} \cdot p)$.
This matches option (C).

AP Precalculus -2.12 Logarithmic Function Manipulation- MCQ Exam Style Questions - Effective Fall 2023