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).

Question

The functions $ g $ and $ h $ are given by:
$$ g(x) = 2 \log a – 3 \log b + \frac{\log a}{2} + \log (100 c) $$
$$ h(x) = \cot x \sec^{2} x – \tan x $$

(A) i. Rewrite $ g(x) $ as a single logarithm with base 10 without negative exponents in any part of the expression. Your result should be simplified in the form $ \log(\text{expression}) $.
ii. Rewrite $ h(x) $ as an expression in which $ \cot x $ appears once and no other trigonometric functions are involved.

(B) The functions $ j $ and $ k $ are given by:
$$ j(x) = 3 \cot^{2} \theta + 3 \csc \theta $$
$$ k(x) = 7 e^{-2x} + 9 $$

i. Solve $ j(x) = -3 $ for values of $ x $ in the interval $ [0, 2\pi) $.
ii. Solve $ k(x) = 86 $ for all values of $ x $ in the domain of $ k $.

(C) Two functions $ a $ and $ b $ are given by:
$$ a(x) = 3 – 3 \csc 3\theta $$
$$ b(x) = -1 + \csc 3\theta $$

Find all the input values in the domain of $ a $ and $ b $ where $ a(x) $ and $ b(x) $ intersect.

Most-appropriate topic codes (CED):

• TOPIC 2.12: Logarithmic Function Manipulation — part (A) i
• TOPIC 3.11: Equivalent Representations of Trigonometric Functions — part (A) ii
• TOPIC 3.12: Trigonometric Equation Solving — part (B) i
• TOPIC 2.13: Exponential and Logarithmic Equations and Inequalities — part (B) ii
• TOPIC 3.12: Trigonometric Equation Solving — part (C)

▶️ Answer/Explanation

Detailed solution

(A) i. Logarithmic Simplification
Start with the given expression:
$ g(x) = 2 \log a + \frac{1}{2} \log a – 3 \log b + \log (100 c) $

Combine the $ \log a $ terms ($ 2 + 0.5 = 2.5 $):
$ \frac{5}{2} \log a – 3 \log b + \log (100 c) $

Use power rules to move coefficients inside the logarithms:
$ \log (a^{5/2}) – \log (b^3) + \log (100 c) $

Combine using product and quotient rules:
$ \log \left( \frac{100 c \cdot a^{5/2}}{b^3} \right) $

Final Answer: $ \log \left( \frac{100 a^{5/2} c}{b^3} \right) $

(A) ii. Trigonometric Simplification
Start with: $ h(x) = \cot x \sec^2 x – \tan x $

Convert to sine and cosine:
$ \left(\frac{\cos x}{\sin x}\right) \cdot \left(\frac{1}{\cos^2 x}\right) – \frac{\sin x}{\cos x} $

Simplify the first term:
$ \frac{1}{\sin x \cos x} – \frac{\sin x}{\cos x} $

Find a common denominator ($ \sin x \cos x $) and simplify:
$ \frac{1 – \sin^2 x}{\sin x \cos x} = \frac{\cos^2 x}{\sin x \cos x} $

Cancel terms:
$ \frac{\cos x}{\sin x} $

Final Answer: $ \cot x $

(B) i. Trigonometric Equation
Set $ j(x) = -3 $:
$ 3 \cot^2 \theta + 3 \csc \theta = -3 $

Divide by 3 and rearrange:
$ \cot^2 \theta + \csc \theta + 1 = 0 $

Substitute identity $ \cot^2 \theta = \csc^2 \theta – 1 $:
$ (\csc^2 \theta – 1) + \csc \theta + 1 = 0 $
$ \csc^2 \theta + \csc \theta = 0 $

Factor:
$ \csc \theta (\csc \theta + 1) = 0 $

Solve $ \csc \theta = -1 $ (since $ \csc \theta \neq 0 $):
$ \sin \theta = -1 $

[Image of unit circle sine values]

In the interval $ [0, 2\pi) $:
Final Answer: $ \theta = \frac{3\pi}{2} $

(B) ii. Exponential Equation
Set $ k(x) = 86 $:
$ 7 e^{-2x} + 9 = 86 $

Subtract 9 and divide by 7:
$ 7 e^{-2x} = 77 \implies e^{-2x} = 11 $

Take the natural log:
$ -2x = \ln 11 $

Final Answer: $ x = -\frac{1}{2} \ln 11 $

(C) Intersection
Set $ a(x) = b(x) $:
$ 3 – 3 \csc 3\theta = -1 + \csc 3\theta $

Rearrange:
$ 4 = 4 \csc 3\theta \implies \csc 3\theta = 1 $

So, $ \sin 3\theta = 1 $. The sine function equals 1 at $ \frac{\pi}{2} $ plus full rotations:
$ 3\theta = \frac{\pi}{2} + 2k\pi $

Final Answer: $ \theta = \frac{\pi}{6} + \frac{2k\pi}{3} $, where $ k \in \mathbb{Z} $

Question

Directions:

Unless otherwise specified, the domain of a function $ f $ is assumed to be the set of all real numbers $ x $ for which $ f(x) $ is a real number. Angle measures for trigonometric functions are assumed to be in radians.
Solutions to equations must be real numbers. Determine the exact value of any expression that can be obtained without a calculator. For example, $ \log_2 8 $, $ \cos(\frac{\pi}{2}) $, and $ \sin^{-1}(1) $ can be evaluated without a calculator.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms, where applicable. For example, $ 2x + 3x $, $ 5^2 \cdot 5^3 $, $ \frac{x^5}{x^2} $, and $ \ln 3 + \ln 5 $ should be rewritten in equivalent forms.
For each part of the question, show the work that leads to your answers.

Part A
The functions $ g $ and $ h $ are given by

$ g(x) = \log_5(4x – 2) $
$ h(x) = \sin^{-1}(8x) $

(i) Solve $ g(x) = 3 $ for values of $ x $ in the domain of $ g $.

(ii) Solve $ h(x) = \frac{\pi}{4} $ for values of $ x $ in the domain of $ h $.

Part B
The functions $ j $ and $ k $ are given by

$ j(x) = (\sec x)(\cot x) $
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $

(i) Rewrite $ j(x) $ as an expression involving $ \sin x $ and no other trigonometric functions.

(ii) Rewrite $ k(x) $ as an expression of the form $ 4^{(ax+b)} $, where $ a $ and $ b $ are constants.

Part C
The function $ m $ is given by

$ m(x) = \sqrt{3}\tan(x + \frac{\pi}{2}) $.

Find all values in the domain of $ m $ that yield an output value of 1.

▶️ Answer/Explanation

Detailed solution

Part A
(i) Solve $ g(x) = 3 $
Start with the given equation:
$ \log_5(4x – 2) = 3 $
Convert the logarithmic equation to exponential form ($ y = \log_b x \iff x = b^y $):
$ 4x – 2 = 5^3 $
Evaluate the exponent:
$ 4x – 2 = 125 $
Add 2 to both sides:
$ 4x = 127 $
Divide by 4:
$ x = \frac{127}{4} $

(ii) Solve $ h(x) = \frac{\pi}{4} $
Start with the given equation:
$ \sin^{-1}(8x) = \frac{\pi}{4} $
Take the sine of both sides to isolate the argument:
$ 8x = \sin\left(\frac{\pi}{4}\right) $
Substitute the exact value of $ \sin\left(\frac{\pi}{4}\right) $:
$ 8x = \frac{\sqrt{2}}{2} $
Divide by 8:
$ x = \frac{\sqrt{2}}{16} $

Part B
(i) Rewrite $ j(x) $
Start with the function definition:
$ j(x) = (\sec x)(\cot x) $
Substitute the reciprocal and quotient identities ($ \sec x = \frac{1}{\cos x} $ and $ \cot x = \frac{\cos x}{\sin x} $):
$ j(x) = \left(\frac{1}{\cos x}\right) \left(\frac{\cos x}{\sin x}\right) $
Cancel the $ \cos x $ terms:
$ j(x) = \frac{1}{\sin x} $

(ii) Rewrite $ k(x) $
Start with the function definition:
$ k(x) = \frac{(16^{3x}) \cdot 4^x}{2} $
To write in the form $ 4^{(ax+b)} $, convert bases 16 and 2 to base 4.
Since $ 16 = 4^2 $ and $ 2 = \sqrt{4} = 4^{1/2} = 4^{0.5} $:
$ k(x) = \frac{(4^2)^{3x} \cdot 4^x}{4^{0.5}} $
Apply the power of a power rule ($ (a^m)^n = a^{mn} $):
$ k(x) = \frac{4^{6x} \cdot 4^x}{4^{0.5}} $
Apply the product rule for exponents ($ a^m \cdot a^n = a^{m+n} $) in the numerator:
$ k(x) = \frac{4^{6x + x}}{4^{0.5}} = \frac{4^{7x}}{4^{0.5}} $
Apply the quotient rule for exponents ($ \frac{a^m}{a^n} = a^{m-n} $):
$ k(x) = 4^{7x – 0.5} $ (or $ 4^{7x – \frac{1}{2}} $)
Thus, $ a = 7 $ and $ b = -0.5 $.

Part C
Find values where $ m(x) = 1 $
Set the function equal to 1:
$ \sqrt{3}\tan\left(x + \frac{\pi}{2}\right) = 1 $
Isolate the tangent function by dividing by $ \sqrt{3} $:
$ \tan\left(x + \frac{\pi}{2}\right) = \frac{1}{\sqrt{3}} $
Determine the reference angle. We know that $ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} $.
Set up the general solution for tangent ($ \theta = \text{ref} + n\pi $):
$ x + \frac{\pi}{2} = \frac{\pi}{6} + n\pi $, where $ n $ is any integer.
Solve for $ x $ by subtracting $ \frac{\pi}{2} $ from both sides:
$ x = \frac{\pi}{6} – \frac{\pi}{2} + n\pi $
Find a common denominator (6) to combine fractions:
$ x = \frac{\pi}{6} – \frac{3\pi}{6} + n\pi $
$ x = -\frac{2\pi}{6} + n\pi $
Simplify the fraction:
$ x = -\frac{\pi}{3} + n\pi $

Question

Directions:

Unless otherwise specified, the domain of a function $ f $ is assumed to be the set of all real numbers $ x $ for which $ f(x) $ is a real number. Angle measures for trigonometric functions are assumed to be in radians.
Solutions to equations must be real numbers. Determine the exact value of any expression that can be obtained without a calculator.
Unless otherwise specified, combine terms using algebraic methods and rules for exponents and logarithms where applicable.
For each part of the question, show the work that leads to your answers.

(A) The functions $ g $ and $ h $ are given by

$ g(x) = 3^{(2x)} \cdot 3^{(x+4)} $
$ h(x) = 2\tan^2 x – 1 $

(i) Solve $ g(x) = 27 $ for values of $ x $ in the domain of $ g $.
(ii) Solve $ h(x) = 5 $ for values of $ x $ in the interval $ [0, 2\pi) $.

(B) The functions $ j $ and $ k $ are given by

$ j(x) = 2\log_{10}(x+3) – \log_{10} x – \log_{10} 3 $
$ k(x) = \frac{(\tan^2 x)(\cot x)}{\csc x} $

(i) Rewrite $ j(x) $ as a single logarithm base 10 without negative exponents in any part of the expression. Your result should be of the form $ \log_{10}(\text{expression}) $.
(ii) Rewrite $ k(x) $ as a fraction involving $ \sec x $ and no other trigonometric functions.

$ m(x) = \frac{2^{(5x+3)}}{\left(2^{(x-2)}\right)^3} $

Find all input values in the domain of $ m $ that yield an output value of $ \frac{1}{16} $.

▶️ Answer/Explanation

Detailed Solution

(A)(i) Solve $ g(x) = 27 $

First, simplify the expression for $ g(x) $ using the property of exponents $ a^m \cdot a^n = a^{m+n} $.
$ g(x) = 3^{2x} \cdot 3^{x+4} = 3^{(2x + x + 4)} = 3^{(3x+4)} $
Set $ g(x) $ equal to 27 and rewrite 27 as a base of 3:
$ 3^{(3x+4)} = 27 $
$ 3^{(3x+4)} = 3^3 $
Since the bases are equal, the exponents must be equal:
$ 3x + 4 = 3 $
$ 3x = 3 – 4 $
$ 3x = -1 $
$ x = -\frac{1}{3} $

(A)(ii) Solve $ h(x) = 5 $ on $ [0, 2\pi) $

Set the expression for $ h(x) $ equal to 5:
$ 2\tan^2 x – 1 = 5 $
Add 1 to both sides:
$ 2\tan^2 x = 6 $
Divide by 2:
$ \tan^2 x = 3 $
Take the square root of both sides:
$ \tan x = \pm\sqrt{3} $
The reference angle for $ \tan \theta = \sqrt{3} $ is $ \frac{\pi}{3} $.
Since we have $ \pm\sqrt{3} $, we must consider solutions in all four quadrants within the interval $ [0, 2\pi) $:
Quadrant I: $ x = \frac{\pi}{3} $
Quadrant II: $ x = \pi – \frac{\pi}{3} = \frac{2\pi}{3} $
Quadrant III: $ x = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $
Quadrant IV: $ x = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} $
Solution set: $ x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} $

(B)(i) Rewrite $ j(x) $ as a single logarithm

Given: $ j(x) = 2\log_{10}(x+3) – \log_{10} x – \log_{10} 3 $
Use the power rule $ n\log a = \log(a^n) $:
$ = \log_{10}((x+3)^2) – \log_{10} x – \log_{10} 3 $
Factor out the negative sign for the last two terms to group them:
$ = \log_{10}((x+3)^2) – (\log_{10} x + \log_{10} 3) $
Use the product rule $ \log a + \log b = \log(ab) $:
$ = \log_{10}((x+3)^2) – \log_{10}(3x) $
Use the quotient rule $ \log a – \log b = \log(\frac{a}{b}) $:
$ = \log_{10}\left(\frac{(x+3)^2}{3x}\right) $

(B)(ii) Rewrite $ k(x) $ involving $ \sec x $

Given: $ k(x) = \frac{(\tan^2 x)(\cot x)}{\csc x} $
First, simplify the numerator using $ \tan x \cdot \cot x = 1 $:
$ (\tan^2 x)(\cot x) = \tan x \cdot (\tan x \cdot \cot x) = \tan x \cdot 1 = \tan x $
Now substitute back into the expression:
$ k(x) = \frac{\tan x}{\csc x} $
Convert to sine and cosine:
$ = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x}} $
$ = \frac{\sin x}{\cos x} \cdot \frac{\sin x}{1} = \frac{\sin^2 x}{\cos x} $
We need the expression in terms of $ \sec x $. Use the identity $ \sin^2 x = 1 – \cos^2 x $:
$ = \frac{1 – \cos^2 x}{\cos x} $
Substitute $ \cos x = \frac{1}{\sec x} $:
$ = \frac{1 – \left(\frac{1}{\sec x}\right)^2}{\frac{1}{\sec x}} $
$ = \frac{1 – \frac{1}{\sec^2 x}}{\frac{1}{\sec x}} $
Find a common denominator for the numerator:
$ = \frac{\frac{\sec^2 x – 1}{\sec^2 x}}{\frac{1}{\sec x}} $
Multiply by the reciprocal of the denominator:
$ = \frac{\sec^2 x – 1}{\sec^2 x} \cdot \frac{\sec x}{1} $
$ = \frac{\sec^2 x – 1}{\sec x} $

(C) Find input values for $ m(x) = \frac{1}{16} $

First, simplify the expression for $ m(x) $:
$ m(x) = \frac{2^{(5x+3)}}{\left(2^{(x-2)}\right)^3} $
Simplify the denominator using the power rule $ (a^m)^n = a^{m \cdot n} $:
$ \left(2^{(x-2)}\right)^3 = 2^{3(x-2)} = 2^{3x-6} $
Now apply the quotient rule $ \frac{a^m}{a^n} = a^{m-n} $:
$ m(x) = \frac{2^{5x+3}}{2^{3x-6}} = 2^{(5x+3) – (3x-6)} $
$ = 2^{5x + 3 – 3x + 6} $
$ = 2^{2x + 9} $
Set $ m(x) = \frac{1}{16} $ and rewrite $ \frac{1}{16} $ as a power of 2:
$ \frac{1}{16} = \frac{1}{2^4} = 2^{-4} $
Equate the simplified $ m(x) $ to $ 2^{-4} $:
$ 2^{2x + 9} = 2^{-4} $
Equate the exponents:
$ 2x + 9 = -4 $
$ 2x = -13 $
$ x = -\frac{13}{2} $ or $ -6.5 $

Question

(A) The function $g$ and $h$ are given by

$g(x) = 3 \ln x – \frac{1}{2} \ln x$

$h(x) = \frac{\sin^2 x – 1}{\cos x}$

(i) Rewrite $g(x)$ as a single natural logarithm without negative exponents in any part of the expression. Your result should be of the form $\ln(\text{expression})$.
(ii) Rewrite $h(x)$ as an expression in which $\cos x$ appears once and no other trigonometric functions are involved.

(B) The functions $j$ and $k$ are given by

$j(x) = 2(\sin x)(\cos x)$

$k(x) = 8e^{(3x)} – e$

(i) Solve $j(x) = 0$ for values of $x$ in the interval $\left[0, \frac{\pi}{2}\right]$.
(ii) Solve $k(x) = 3e$ for values of $x$ in the domain of $k$.

$m(x) = \cos(2x) + 4$

Find all input values in the domain of $m$ that yield an output of $\frac{9}{2}$.

▶️ Answer/Explanation

Detailed solution

(A)(i) Rewrite $g(x)$

The function is given by $g(x) = 3 \ln x – \frac{1}{2} \ln x$.
Combine the like terms:
$g(x) = \left(3 – \frac{1}{2}\right) \ln x$
$g(x) = \frac{5}{2} \ln x$
Apply the power property of logarithms, $a \ln b = \ln(b^a)$:
$g(x) = \ln\left(x^{5/2}\right)$

(A)(ii) Rewrite $h(x)$

The function is given by $h(x) = \frac{\sin^2 x – 1}{\cos x}$.
Recall the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$.
Rearrange the identity to isolate the numerator expression: $\sin^2 x – 1 = -\cos^2 x$.
Substitute this into the function:
$h(x) = \frac{-\cos^2 x}{\cos x}$
Simplify the expression by canceling one $\cos x$ term:
$h(x) = -\cos x$

(B)(i) Solve $j(x) = 0$

Set the function equal to zero: $2(\sin x)(\cos x) = 0$.
Divide both sides by 2:
$\sin x \cos x = 0$
By the zero product property, either $\sin x = 0$ or $\cos x = 0$.
Case 1: $\sin x = 0$. In the interval $\left[0, \frac{\pi}{2}\right]$, $x = 0$.
Case 2: $\cos x = 0$. In the interval $\left[0, \frac{\pi}{2}\right]$, $x = \frac{\pi}{2}$.
The solutions are $x = 0$ and $x = \frac{\pi}{2}$.

(B)(ii) Solve $k(x) = 3e$

Set the function equal to $3e$:
$8e^{(3x)} – e = 3e$
Add $e$ to both sides:
$8e^{(3x)} = 4e$
Divide both sides by 8:
$e^{(3x)} = \frac{4e}{8}$
$e^{(3x)} = \frac{e}{2}$
Take the natural logarithm ($\ln$) of both sides:
$\ln\left(e^{3x}\right) = \ln\left(\frac{e}{2}\right)$
Use logarithm properties to simplify ($\ln(e^a) = a$ and $\ln(a/b) = \ln a – \ln b$):
$3x = \ln e – \ln 2$
Since $\ln e = 1$:
$3x = 1 – \ln 2$
Divide by 3:
$x = \frac{1 – \ln 2}{3}$

(C) Find input values for $m(x) = \frac{9}{2}$

Set the function equal to $\frac{9}{2}$:
$\cos(2x) + 4 = \frac{9}{2}$
Subtract 4 from both sides (note that $4 = \frac{8}{2}$):
$\cos(2x) = \frac{9}{2} – \frac{8}{2}$
$\cos(2x) = \frac{1}{2}$
The reference angle for cosine equal to $\frac{1}{2}$ is $\frac{\pi}{3}$.
The general solution for $2x$ is:
$2x = \frac{\pi}{3} + 2\pi n$ or $2x = -\frac{\pi}{3} + 2\pi n$ (where $n$ is an integer).
Solve for $x$ by dividing by 2:
$x = \frac{\pi}{6} + \pi n$ or $x = -\frac{\pi}{6} + \pi n$
Combining these, the input values are $x = \pm \frac{\pi}{6} + \pi n$ for any integer $n$.

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