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.

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