AP Precalculus -2.9 Logarithmic Expressions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.9 Logarithmic Expressions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.9 Logarithmic Expressions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ h $ is given by $ h(x) = 5 \cdot 3^{(-x/2)} $. What is the value of $ h(1) $?

(A) $ -5\sqrt{3} $
(B) $ \frac{1}{\sqrt{15}} $
(C) $ \frac{5}{9} $
(D) $ \frac{5}{\sqrt{3}} $

▶️ Answer/Explanation

Detailed solution

$ h(1) = 5 \cdot 3^{(-1/2)} = 5 \cdot \frac{1}{3^{1/2}} = \frac{5}{\sqrt{3}} $.
✅ Answer: (D)

Question

The function $f$ is given by $f(x)=\log_{2}x$. What input value in the domain of $f$ yields an output value of $4$?

(A) $32$
(B) $16$
(C) $2$
(D) $\frac{1}{2}$

▶️ Answer/Explanation

Detailed solution

1. Set up the equation:
Set $f(x) = 4$:
$\log_{2}x = 4$

2. Convert to exponential form:
$x = 2^4$

3. Calculate:
$x = 16$

✅ Answer: (B)

Question

Which of the following expressions is equivalent to $\ln \left(\frac{5e^2}{\sqrt{x}}\right)$, where $x$ is a positive constant?

(A) $2 + \ln 5 – \frac{1}{2}\ln x$

(B) $2 \ln 5 – \frac{1}{2}\ln x$

(D) $2 \ln 5 – 2 \ln x$

▶️ Answer/Explanation

Detailed solution

To simplify the expression $\ln \left(\frac{5e^2}{\sqrt{x}}\right)$, apply the laws of logarithms step-by-step.
1. Use the quotient rule $\ln(\frac{a}{b}) = \ln(a) – \ln(b)$:
$\quad \ln(5e^2) – \ln(\sqrt{x})$
2. Use the product rule $\ln(ab) = \ln(a) + \ln(b)$ on the first term:
$\quad \ln(5) + \ln(e^2) – \ln(\sqrt{x})$
3. Simplify $\ln(e^2) = 2$ (since $\ln e = 1$) and rewrite $\sqrt{x}$ as $x^{1/2}$:
$\quad \ln(5) + 2 – \ln(x^{1/2})$
4. Use the power rule $\ln(a^k) = k \ln(a)$ on the last term:
$\quad 2 + \ln(5) – \frac{1}{2}\ln(x)$
This matches option (A).

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) $x \log_{3}\left(\frac{27}{9}\right)$

(D) $\frac{x \log_{3} 27}{\log_{3} 9}$

▶️ Answer/Explanation

Detailed solution

First, apply the power rule of logarithms: $\log_{9}(27^{x}) = x \log_{9} 27$.
Apply the change-of-base formula $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ to rewrite the expression.
Choose base $3$ because both $27$ and $9$ are powers of $3$.
This gives the equivalent form: $\frac{x \log_{3} 27}{\log_{3} 9}$.
Since $\log_{3} 27 = 3$ and $\log_{3} 9 = 2$, the expression becomes $\frac{3}{2}x$.
This matches choice (D), allowing for simple mental calculation of the rational multiple.
Correct Option: (D)

Question

If $m = \log_{3} 81$, which of the following is also true?

(A) $3m = 81$

(B) $3^m = 81$

(D) $\sqrt[3]{81} = m$

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
The given equation is $m = \log_{3} 81$.
By the definition of a logarithm, $\log_{b} x = y$ is equivalent to $b^y = x$.
In this case, the base $b$ is $3$, the result $y$ is $m$, and the argument $x$ is $81$.
Converting the logarithmic form to exponential form gives $3^m = 81$.
Since $3^4 = 81$, it also follows that $m = 4$.
Thus, the expression $3^m = 81$ is the true statement.

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

The given function is $g(x) = a \log_{b} c$.
According to the power property of logarithms, $n \log_{x} y = \log_{x}(y^{n})$.
Applying this property, the coefficient $a$ becomes the exponent of the argument $c$.
This results in the equivalent expression $\log_{b}(c^{a})$.
Option (A) correctly matches this logarithmic identity.
Options (B), (C), and (D) violate standard logarithmic laws.
Therefore, the correct representation is option (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) $\log_{10} k + \log_{10} z – 2\log_{10} w$

(D) $\log_{10} k – \log_{10} z + 2\log_{10} w$

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
Apply the quotient rule: $\log_{10}\left(\frac{kz}{w^2}\right) = \log_{10}(kz) – \log_{10}(w^2)$.
Apply the product rule: $\log_{10}(kz) = \log_{10} k + \log_{10} z$.
Apply the power rule: $\log_{10}(w^2) = 2\log_{10} w$.
Combine the terms: $\log_{10} k + \log_{10} z – 2\log_{10} w$.
This matches the expression provided in option (B).

Question

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

(A) $\log_{3} 5 + \log_{3} x$

(B) $\log_{3} 5 \cdot \log_{3} x$

(D) $\frac{\log_{3} x}{\log_{3} 5}$

▶️ Answer/Explanation

Detailed solution

The correct option is (C).
This problem uses the Power Property of Logarithms.
The rule states that $\log_{b}(m^{n}) = n \log_{b}(m)$.
In the expression $\log_{3}(x^{5})$, the base $b$ is $3$, the argument $m$ is $x$, and the exponent $n$ is $5$.
By moving the exponent to the front as a multiplier, we get $5 \log_{3} x$.
Therefore, $\log_{3}(x^{5})$ is equivalent to $5 \log_{3} x$.

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) $\ln(x^2 y^3)$

(D) $\ln\left(\frac{2x}{3y}\right)$

▶️ Answer/Explanation

Detailed solution

The given expression is $2 \ln x – 3 \ln y$.
Apply the Power Property: $n \ln a = \ln(a^n)$ to get $\ln(x^2) – \ln(y^3)$.
Apply the Quotient Property: $\ln a – \ln b = \ln\left(\frac{a}{b}\right)$.
Combine the terms: $\ln\left(\frac{x^2}{y^3}\right)$.
This matches the expression in option (A).
Correct Option: (A)

Question

Which of the following expressions is equivalent to $\ln\left(\frac{2e^4}{x^3}\right)$, where $x$ is a positive constant?

(A) $4 + \ln 2 – 3\ln x$

(B) $4 + \ln 2 + 3\ln x$

(D) $\ln 2 + \ln 4 – 3\ln x$

▶️ Answer/Explanation

Detailed solution

Start with the given expression: $\ln\left(\frac{2e^4}{x^3}\right)$.
Apply the quotient property of logarithms, $\ln(\frac{a}{b}) = \ln a – \ln b$, to separate the numerator and denominator: $\ln(2e^4) – \ln(x^3)$.
Apply the product property, $\ln(ab) = \ln a + \ln b$, to the first term: $\ln 2 + \ln(e^4) – \ln(x^3)$.
Use the power property, $\ln(a^k) = k \ln a$, to move the exponents to the front: $\ln 2 + 4\ln e – 3\ln x$.
Recall that the natural logarithm of $e$ is 1 ($\ln e = 1$), so substitute this value: $\ln 2 + 4(1) – 3\ln x$.
Rearrange the terms to match the format of the options: $4 + \ln 2 – 3\ln x$.
This corresponds to option (A).

Question

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

(A) $\ln \left( \frac{2w}{y \cdot 3z} \right)$

(B) $\ln \left( \frac{w^2}{y \cdot z^3} \right)$

(D) $\ln (w^2 \cdot y \cdot z^3)$

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
Apply the power rule: $n \ln x = \ln x^n$, giving $\ln w^2 – \ln y – \ln z^3$.
Factor out the negative sign: $\ln w^2 – (\ln y + \ln z^3)$.
Apply the product rule: $\ln y + \ln z^3 = \ln(y \cdot z^3)$.
Apply the quotient rule: $\ln w^2 – \ln(y \cdot z^3) = \ln \left( \frac{w^2}{y \cdot z^3} \right)$.
This matches the simplified expression shown in the handwritten note.
Therefore, the expression is condensed into a single logarithm.

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

(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 -2.9 Logarithmic Expressions- MCQ Exam Style Questions - Effective Fall 2023