AP Precalculus -2.13 Exponential and Logarithmic Equations and Inequalities- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -2.13 Exponential and Logarithmic Equations and Inequalities- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -2.13 Exponential and Logarithmic Equations and Inequalities- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The range of function $ f $ is the positive real numbers. The function $ g $ is given by $ g(x) = \ln(f(x)) $. Solutions to which of the following equations are useful in solving $ g(x) = 2 $?

(A) $ f(x) = 2 $
(B) $ f(x) = e^2 $
(C) $ f(x) = 10^2 $
(D) $ f(x) = \frac{2}{\ln x} $

▶️ Answer/Explanation

Detailed solution

$ g(x) = 2 $ ⇒ $ \ln(f(x)) = 2 $ ⇒ $ f(x) = e^2 $ by definition of natural log.
✅ Answer: (B)

Question

What are all values of $ x $ for which $ \ln(x^3) – \ln x = 4 $?

(A) $ x = -2 $ and $ x = 2 $
(B) $ x = -e^2 $ and $ x = e^2 $
(C) $ x = e^2 $ only
(D) $ x = e^4 $

▶️ Answer/Explanation

Detailed solution

Combine logs: $ \ln\left(\frac{x^3}{x}\right) = \ln(x^2) = 4 $.
So $ x^2 = e^4 $ ⇒ $ x = \pm e^2 $.
Domain of original: $ \ln(x^3) $ requires $ x>0 $, and $ \ln x $ requires $ x>0 $, so $ x>0 $.
Thus $ x = e^2 $ only.
✅ Answer: (C)

Question

Consider the functions $ f $ and $ g $ given by $ f(x) = \log_{10}(x – 1) + \log_{10}(x + 3) $ and $ g(x) = \log_{10}(x + 9) $. In the $ xy $-plane, what are all $ x $-coordinates of the points of intersection of the graphs of $ f $ and $ g $?

(A) $ x = 3 $ only
(B) $ x = 7 $
(C) $ x = -4 $ and $ x = 3 $
(D) $ x = -7 $ and $ x = -4 $

▶️ Answer/Explanation

Detailed solution

Combine $ f $: $ f(x) = \log_{10}[(x-1)(x+3)] $.
Set $ f(x) = g(x) $: $ \log_{10}[(x-1)(x+3)] = \log_{10}(x+9) $ ⇒ $ (x-1)(x+3) = x+9 $.
Expand: $ x^2 + 2x – 3 = x + 9 $ ⇒ $ x^2 + x – 12 = 0 $ ⇒ $ (x+4)(x-3) = 0 $ ⇒ $ x = -4 $ or $ x = 3 $.
Domain of $ f $: $ x-1 > 0 $ ⇒ $ x>1 $ and $ x+3>0 $ ⇒ $ x>-3 $, so $ x>1 $.
Thus $ x = -4 $ is extraneous; only $ x=3 $.
✅ Answer: (A)

Question

The function $ g $ is given by $ g(x) = \ln(3x + 1) – \ln(x^2 + x – 2) $. What are all values of $ x $ for which $ g(x) < 0 $?

(A) $ (-\infty, -1) $ and $ (3, \infty) $
(B) $ (-1, 3) $
(C) $ (1, 3) $ only
(D) $ (3, \infty) $ only

▶️ Answer/Explanation

Detailed solution

$ g(x) < 0 $ ⇒ $ \ln\frac{3x+1}{x^2 + x – 2} < 0 $ ⇒ $ \frac{3x+1}{x^2 + x – 2} < 1 $, given the fraction is positive (domain).
Domain: $ 3x+1 > 0 $ ⇒ $ x > -\frac{1}{3} $, and $ x^2 + x – 2 > 0 $ ⇒ $ (x+2)(x-1) > 0 $ ⇒ $ x < -2 $ or $ x > 1 $.
Intersection: $ x > 1 $.
Within $ x>1 $, solve $ \frac{3x+1}{x^2 + x – 2} < 1 $.
Since denominator > 0, multiply: $ 3x+1 < x^2 + x – 2 $ ⇒ $ 0 < x^2 – 2x – 3 $ ⇒ $ (x-3)(x+1) > 0 $ ⇒ $ x < -1 $ or $ x > 3 $.
Combine with $ x>1 $: $ x > 3 $.
✅ Answer: (D)

Question

A decibel (dB) is a unit of measure for loudness of sound. The decibel scale is based on sound intensity $ N $, in watts per square meter. A decibel value is given by the function $ d(N) = 10 \log_{10} \left( \frac{N}{10^{-12}} \right) $. Which of the following gives all intensities $ N $, in watts per square meter, for which the decibel value is greater than 140 decibels?

(A) $ N > 14 \cdot 10^{-12} $
(B) $ N > 2 $
(C) $ N > 100 $
(D) $ N > 10^{26} $

▶️ Answer/Explanation

Detailed solution

$ d(N) > 140 $ ⇒ $ 10 \log_{10} \left( \frac{N}{10^{-12}} \right) > 140 $
Divide by 10: $ \log_{10} \left( \frac{N}{10^{-12}} \right) > 14 $
Convert: $ \frac{N}{10^{-12}} > 10^{14} $
Multiply: $ N > 10^{14} \cdot 10^{-12} = 10^{2} = 100 $.
✅ Answer: (C)

Question

Consider the functions $ f $ and $ g $ given by $ f(x) = \log_{10}(x-1) + \log_{10}(x+3) $ and $ g(x) = \log_{10}(x+9) $. In the xy-plane, what are all $ x $-coordinates of the points of intersection of the graphs of $ f $ and $ g $?

(A) $ x = 3 $ only
(B) $ x = 7 $
(C) $ x = -4 $ and $ x = 3 $
(D) $ x = -7 $ and $ x = -4 $

▶️ Answer/Explanation

Detailed solution

Combine logs for $ f $:
\[ f(x) = \log_{10}[(x-1)(x+3)] = \log_{10}(x^2 + 2x – 3) \]
Set $ f(x) = g(x) $:
\[ \log_{10}(x^2 + 2x – 3) = \log_{10}(x+9) \]
Equate arguments (since logs are same base and function is 1–1):
\[ x^2 + 2x – 3 = x + 9 \]
\[ x^2 + x – 12 = 0 \]
\[ (x+4)(x-3) = 0 \]
$ x = -4 $ or $ x = 3 $.
Check domain: for $ f $, need $ x-1 > 0 $ and $ x+3 > 0 $ ⇒ $ x > 1 $. So $ x = -4 $ is extraneous.
Only $ x = 3 $ works.
✅ Answer: (A)

Question

For which values of $x$ does $\ln(-3x – 7) – \ln 7 = 5$?

a. $x = \frac{7e^5+7}{-3}$
b. $x = \frac{-3e+1}{3}$
c. $x = 167$
d. $x = \frac{-e^2-24}{6}$

▶️ Answer/Explanation

Detailed solution

The correct option is a.
Start with the equation: $\ln(-3x – 7) – \ln 7 = 5$.
Apply the quotient rule for logarithms: $\ln\left(\frac{-3x – 7}{7}\right) = 5$.
Rewrite in exponential form: $\frac{-3x – 7}{7} = e^5$.
Multiply both sides by $7$ to get: $-3x – 7 = 7e^5$.
Add $7$ to both sides: $-3x = 7e^5 + 7$.
Solve for $x$ by dividing by $-3$: $x = \frac{7e^5 + 7}{-3}$.
Check domain: $-3x-7 > 0$ is satisfied by this value.

Question

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

(D) 357,143

▶️ Answer/Explanation

Detailed solution

Given the logistic function $S(t) = \frac{500,000}{1+0.4e^{kt}}$ and the condition $S(4) = 300,000$:
First, substitute $t=4$ to solve for the exponential term:
$300,000 = \frac{500,000}{1+0.4e^{4k}} \Rightarrow 1+0.4e^{4k} = \frac{5}{3} \Rightarrow 0.4e^{4k} = \frac{2}{3}$
Solving for $e^{4k}$, we get $e^{4k} = \frac{2}{3} \div 0.4 = \frac{5}{3}$.

Next, find $S(12)$. Note that $e^{12k} = (e^{4k})^3$:
$e^{12k} = (\frac{5}{3})^3 = \frac{125}{27}$
Substitute this back into the equation for $S(12)$:
$S(12) = \frac{500,000}{1+0.4(\frac{125}{27})} = \frac{500,000}{1+\frac{2}{5}(\frac{125}{27})} = \frac{500,000}{1+\frac{50}{27}}$

Simplify the final expression:
$S(12) = \frac{500,000}{\frac{77}{27}} = \frac{500,000 \times 27}{77} \approx 175,324.67$
Rounding to the nearest integer, the value is 175,325.

Correct Option: (A)

Question

The function $h$ is given by $h(x) = 8 \cdot 2^{x}$. For which of the following values of $x$ is $h(x) = 256$ ?

(A) $x = 2$
(B) $x = 5$
(C) $x = 8$
(D) $x = 16$

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $8 \cdot 2^{x} = 256$.
Divide both sides by $8$ to isolate the exponential term: $2^{x} = \frac{256}{8}$.
Simplify the division: $2^{x} = 32$.
Express $32$ as a power of $2$: $32 = 2^{5}$.
Equate the exponents since the bases are the same: $x = 5$.
Therefore, the correct option is (B).

Question

Consider the functions $g$ and $h$ given by $g(x) = 4^{x}$ and $h(x) = 16^{x+2}$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $g$ and $h$?

(A) $-4$

(B) $-2$

(D) $2$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set the functions equal: $g(x) = h(x)$.
Substitute the expressions: $4^{x} = 16^{x+2}$.
Rewrite $16$ as a power of $4$: $4^{x} = (4^{2})^{x+2}$.
Apply the power of a power rule: $4^{x} = 4^{2(x+2)}$.
Set the exponents equal to each other: $x = 2(x+2)$.
Distribute the $2$: $x = 2x + 4$.
Subtract $2x$ from both sides: $-x = 4$.
Solve for $x$: $x = -4$.
The correct option is (A).

Question

The range of function $f$ is the positive real numbers. The function $g$ is given by $g(x) = \ln(f(x))$. Solutions to which of the following equations are useful in solving $g(x) = 2$?

(A) $f(x) = 2$
(B) $f(x) = e^2$
(C) $f(x) = 10^2
(D) $f(x) = \frac{2}{\ln x}$

▶️ Answer/Explanation

Detailed solution

The given equation is $g(x) = 2$.
Substitute the definition of $g(x)$ into the equation: $\ln(f(x)) = 2$.
Recall that the natural logarithm $\ln(y)$ has a base of $e$.
Rewrite the logarithmic equation in its equivalent exponential form.
If $\log_{b}(a) = c$, then $b^c = a$; here, $e^2 = f(x)$.
Therefore, solving $f(x) = e^2$ provides the values of $x$ that satisfy $g(x) = 2$.
The correct option is (B).

Question

What are all values of $x$ for which $\ln(x^3) – \ln x = 4$?

(A) $x = -2$ and $x = 2$
(B) $x = -e^2$ and $x = e^2$
(C) $x = e^2$ only
(D) $x = e^4$

▶️ Answer/Explanation

Detailed solution

The domain of $\ln x$ requires that $x > 0$.
Using the quotient rule for logarithms: $\ln\left(\frac{x^3}{x}\right) = 4$.
Simplify the fraction: $\ln(x^2) = 4$.
Convert the logarithmic equation to exponential form: $x^2 = e^4$.
Solve for $x$ by taking the square root: $x = \pm \sqrt{e^4} = \pm e^2$.
Checking the domain, $x$ must be positive, so $x = -e^2$ is extraneous.
Therefore, the only valid solution is $x = e^2$.
The correct option is (C).

Question

Consider the functions $f$ and $g$ given by $f(x) = \log_{10}(x – 1) + \log_{10}(x + 3)$ and $g(x) = \log_{10}(x + 9)$. In the $xy$-plane, what are all $x$-coordinates of the points of intersection of the graphs of $f$ and $g$?

(A) $x = 3$ only
(B) $x = 7$
(C) $x = -4$ and $x = 3$
(D) $x = -7$ and $x = -4$

▶️ Answer/Explanation

Detailed solution

Set the functions equal: $\log_{10}(x – 1) + \log_{10}(x + 3) = \log_{10}(x + 9)$.
Apply the product property: $\log_{10}((x – 1)(x + 3)) = \log_{10}(x + 9)$.
Remove logs and expand: $x^2 + 2x – 3 = x + 9$.
Rearrange into a quadratic equation: $x^2 + x – 12 = 0$.
Factor the quadratic: $(x + 4)(x – 3) = 0$, giving $x = -4$ and $x = 3$.
Check domain constraints: For $f(x)$, $x – 1 > 0$, so $x > 1$.
Reject $x = -4$ as it is outside the domain; thus, $x = 3$ only is the valid solution.
The correct option is (A).

Question

The function $g$ is given by $g(x) = \ln(3x + 1) – \ln(x^2 + x – 2)$. What are all values of $x$ for which $g(x) < 0$?

(A) $(-\infty, -1)$ and $(3, \infty)$

(B) $(-1, 3)$

(D) $(3, \infty)$ only

▶️ Answer/Explanation

Detailed solution

First, identify the domain: $3x + 1 > 0 \implies x > -\frac{1}{3}$ and $x^2 + x – 2 > 0 \implies (x+2)(x-1) > 0$.
The intersection of $x > -\frac{1}{3}$ and $(x < -2 \text{ or } x > 1)$ gives the domain $x > 1$.
Set $g(x) < 0$: $\ln(3x + 1) – \ln(x^2 + x – 2) < 0 \implies \ln(3x + 1) < \ln(x^2 + x – 2)$.
Since $\ln$ is increasing, $3x + 1 < x^2 + x – 2$.
Rearrange to $x^2 – 2x – 3 > 0$, which factors as $(x – 3)(x + 1) > 0$.
This inequality holds for $x < -1$ or $x > 3$.
Combining $x > 1$ (domain) and $(x < -1 \text{ or } x > 3)$ yields $x > 3$.
The correct option is (D).

Question

A decibel (dB) is a unit of measure for loudness of sound. The decibel scale is based in sound intensity $N$, in watts per square meter. A decibel value is given by the function $d$, where $d(N) = 10 \log_{10} \left( \frac{N}{10^{-12}} \right)$. Which of the following gives all intensities $N$, in watts per square meter, for which the decibel value is greater than $140$ decibels?

(A) $N > 14 \cdot 10^{-12}$
(B) $N > 2$
(C) $N > 100$
(D) $N > 10^{26}$

▶️ Answer/Explanation

Detailed solution

Set the decibel function $d(N) > 140$.
Substitute the formula: $10 \log_{10} \left( \frac{N}{10^{-12}} \right) > 140$.
Divide both sides by $10$ to get $\log_{10} \left( \frac{N}{10^{-12}} \right) > 14$.
Rewrite the logarithmic inequality in exponential form: $\frac{N}{10^{-12}} > 10^{14}$.
Multiply both sides by $10^{-12}$ to isolate $N$.
$N > 10^{14} \cdot 10^{-12}$.
Simplify the exponents by adding them: $N > 10^{14 + (-12)}$.
The final result is $N > 10^2$, which is $N > 100$.
Correct Option: (C)

Question

What are all values of $ x $ for which $ \ln(3x) + \ln 2 = 4 $?

(A) $ x = \frac{2}{3} $

(B) $ x = \frac{e^4}{6} $

(D) $ x = \frac{e^4 – 2}{3} $

▶️ Answer/Explanation

Detailed solution

The correct option is (B).

Step 1: Start with the given equation: $ \ln(3x) + \ln 2 = 4 $.
Step 2: Apply the product property of logarithms, $ \ln a + \ln b = \ln(ab) $, to combine the terms on the left side:
$ \ln(3x \cdot 2) = 4 \implies \ln(6x) = 4 $
Step 3: Convert the logarithmic equation to its exponential form using the definition $ \ln y = k \iff y = e^k $:
$ 6x = e^4 $
Step 4: Solve for $ x $ by dividing both sides by 6:
$ x = \frac{e^4}{6} $

Question

What are all values of $x$ for which $\log_{7}(x-1) + 2 = \log_{7}(4x+6)$?

(A) $x = -\frac{5}{3}$

(B) $x = -4$

(D) there are no such values of $x$

▶️ Answer/Explanation

Detailed solution

Express the constant as a logarithm: $2 = \log_{7}(7^2) = \log_{7}(49)$.
Combine terms using the product rule: $\log_{7}(49(x-1)) = \log_{7}(4x+6)$.
Equate the arguments of the logarithms: $49x – 49 = 4x + 6$.
Solve the linear equation: $45x = 55$, which simplifies to $x = \frac{11}{9}$.
Check constraints: For $\log_{7}(x-1)$, we require $x – 1 > 0 \Rightarrow x > 1$.
Verify the solution: Since $\frac{11}{9} \approx 1.22$, which is $> 1$, the value is valid.
Final Answer: (C)

Question

Consider the functions $g$ and $h$ given by $g(x) = 4^{x+1}$ and $h(x) = 2^{3x}$. Let $f$ be the function given by $f(x) = \frac{g(x)}{h(x)}$. In the $xy$-plane, what is the $x$-coordinate of the point where $f(x) = 1$?

(A) $\frac{1}{5}$
(B) $\frac{1}{2}$
(C) $1$
(D) $2$

▶️ Answer/Explanation

Detailed solution

Set the function $f(x) = 1$, which implies $\frac{g(x)}{h(x)} = 1$, or $g(x) = h(x)$.
Substitute the given expressions: $4^{x+1} = 2^{3x}$.
Rewrite the base $4$ as $2^2$: $(2^2)^{x+1} = 2^{3x}$.
Apply the power of a power rule: $2^{2(x+1)} = 2^{3x}$.
Equate the exponents since the bases are equal: $2(x+1) = 3x$.
Expand the left side: $2x + 2 = 3x$.
Subtract $2x$ from both sides to solve for $x$: $x = 2$.
The $x$-coordinate is $2$, which corresponds to option (D).

Question

The function $f$ is given by $f(x) = 3 + 2e^{-x}$. For which of the following values of $x$ is $f(x) = 11$?

(A) $x = 1 + \ln 4$
(B) $x = \dfrac{-\ln 8}{2}$
(C) $x = -\ln 4$
(D) $x = \dfrac{3 – \ln 11}{2}$

▶️ Answer/Explanation

Detailed solution

Set the function equal to $11$: $3 + 2e^{-x} = 11$.
Subtract $3$ from both sides to isolate the exponential term: $2e^{-x} = 8$.
Divide both sides by $2$ to simplify: $e^{-x} = 4$.
Take the natural logarithm ($\ln$) of both sides: $-x = \ln 4$.
Multiply by $-1$ to solve for $x$: $x = -\ln 4$.
Match the result with the given options: **Correct Option (C)**.

Question

Consider the functions $f$ and $g$ given by $f(x) = \log_{10}(x+3) – \log_{10}(x)$ and $g(x) = 1$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $f$ and $g$?

(A) $-\frac{10}{3}$

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

(D) $3$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set $f(x) = g(x)$, which gives $\log_{10}(x+3) – \log_{10}(x) = 1$.
Use the quotient rule for logarithms: $\log_{10}\left(\frac{x+3}{x}\right) = 1$.
Rewrite the logarithmic equation in exponential form: $\frac{x+3}{x} = 10^1$.
Multiply both sides by $x$ to get $x + 3 = 10x$.
Subtract $x$ from both sides: $3 = 9x$.
Solve for $x$ by dividing: $x = \frac{3}{9} = \frac{1}{3}$.
Since $x = \frac{1}{3}$ is positive, it satisfies the domain $x > 0$.
The correct option is (B).

Question

Consider the functions $f$ and $g$ given by $f(x) = \ln(2x + 1)$ and $g(x) = 2\ln 3$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $f$ and $g$?

(A) $-5$
(B) $\frac{5}{2}$
(C) $\frac{7}{2}$
(D) $4$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set the functions equal: $f(x) = g(x)$.
Substitute the given expressions: $\ln(2x + 1) = 2\ln 3$.
Use the power property of logarithms: $\ln(2x + 1) = \ln(3^2)$.
Simplify the right side: $\ln(2x + 1) = \ln 9$.
Remove the logarithms: $2x + 1 = 9$.
Solve for $x$: $2x = 8$.
Final result: $x = 4$.
The correct option is (D).

Question

Consider the functions $f$ and $g$ given by $f(x) = \log_{2}(3x – 1) – \log_{2}(x + 1)$ and $g(x) = \log_{2}(2)$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $f$ and $g$?

(A) $-5$
(B) $2$
(C) $3$
(D) there is no such value of $x$

▶️ Answer/Explanation

Detailed solution

Set the functions equal: $\log_{2}(3x – 1) – \log_{2}(x + 1) = \log_{2}(2)$.
Use the quotient rule for logarithms: $\log_{2}\left(\frac{3x – 1}{x + 1}\right) = \log_{2}(2)$.
Equate the arguments: $\frac{3x – 1}{x + 1} = 2$.
Multiply both sides by $(x + 1)$: $3x – 1 = 2(x + 1)$.
Expand and simplify: $3x – 1 = 2x + 2$.
Solve for $x$: $x = 3$.
Check domain: $3(3)-1 > 0$ and $3+1 > 0$, so $x = 3$ is valid.
The correct option is (C).

Question

Consider the functions $f$ and $g$ given by $f(x) = \log_{3} x – \log_{3} 4$ and $g(x) = 2$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $f$ and $g$?

(A) $\frac{9}{4}$
(B) $8$
(C) $32$
(D) $36$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set the functions equal: $f(x) = g(x)$.
Substitute the given expressions: $\log_{3} x – \log_{3} 4 = 2$.
Apply the quotient rule for logarithms: $\log_{3} \left( \frac{x}{4} \right) = 2$.
Rewrite the logarithmic equation in exponential form: $\frac{x}{4} = 3^{2}$.
Simplify the exponent: $\frac{x}{4} = 9$.
Multiply both sides by $4$ to solve for $x$: $x = 36$.
The $x$-coordinate of the point of intersection is 36.
Therefore, the correct option is (D).

Question

Consider the functions $f$ and $g$ given by $f(x) = \log_{10}(x+3) + \log_{10}(x-5)$ and $g(x) = \log_{10}(3x-1)$. In the $xy$-plane, what are all $x$-coordinates of the points of intersection of the graphs of $f$ and $g$?

(A) $-2$ and $7$
(B) $-2$ only
(C) $2$ only
(D) $7$ only

▶️ Answer/Explanation

Detailed solution

Set the functions equal to find the intersection: $\log_{10}(x+3) + \log_{10}(x-5) = \log_{10}(3x-1)$.
Apply the product rule for logarithms: $\log_{10}((x+3)(x-5)) = \log_{10}(3x-1)$.
Equate the arguments: $(x+3)(x-5) = 3x-1$.
Expand and simplify to a quadratic: $x^2 – 2x – 15 = 3x – 1$, which becomes $x^2 – 5x – 14 = 0$.
Factor the quadratic equation: $(x-7)(x+2) = 0$, yielding potential solutions $x = 7$ and $x = -2$.
Check the domain: for $f(x)$, $x+3 > 0$ and $x-5 > 0$, requiring $x > 5$.
Since $x = -2$ is not in the domain ($x-5$ would be negative), only $x = 7$ is valid.
Correct Option: (D) $7$ only

Question

Consider the functions $f$ and $g$ given by $f(x) = \ln(2x+3) + \ln(x-1)$ and $g(x) = \ln(x^2 + 6x – 9)$. In the $xy$-plane, what are all of the points of intersection of the graphs of $f$ and $g$?

(A) $2$ and $3$
(B) $6$ only
(C) $6$ and $-1$
(D) $\frac{-3 + \sqrt{53}}{2}$

▶️ Answer/Explanation

Detailed solution

Set $f(x) = g(x)$, so $\ln((2x+3)(x-1)) = \ln(x^2 + 6x – 9)$.
Simplify the arguments: $2x^2 + x – 3 = x^2 + 6x – 9$.
Rearrange into a quadratic equation: $x^2 – 5x + 6 = 0$.
Factor the quadratic to find potential solutions: $(x-2)(x-3) = 0$, giving $x = 2$ and $x = 3$.
Check domain constraints: $2x+3 > 0$, $x-1 > 0$, and $x^2+6x-9 > 0$.
For $x=2$: $2(2)+3=7$, $2-1=1$, and $4+12-9=7$ (all positive).
For $x=3$: $2(3)+3=9$, $3-1=2$, and $9+18-9=18$ (all positive).
Both $x=2$ and $x=3$ are valid points of intersection.
The correct option is (A).

Question

What are all values of $x$ for which $\log_{2}(x^{2}) – \log_{2} 4 = 3$?

(A) $x = -\sqrt{2}$ and $x = \sqrt{2}$

(B) $x = -\sqrt{7}$ and $x = \sqrt{7}$

(D) $x = -4\sqrt{2}$ and $x = 4\sqrt{2}$

▶️ Answer/Explanation

Detailed solution

The correct option is (D).
Given: $\log_{2}(x^{2}) – \log_{2} 4 = 3$
Use quotient rule: $\log_{2}\left(\frac{x^{2}}{4}\right) = 3$
Convert to exponential form: $\frac{x^{2}}{4} = 2^{3}$
Simplify the power: $\frac{x^{2}}{4} = 8$
Multiply both sides by 4: $x^{2} = 32$
Solve for $x$: $x = \pm\sqrt{32}$
Simplify the radical: $x = \pm 4\sqrt{2}$

Question

What are all values of $x$ for which $\ln(3x) + \ln 2 = 4$?

(A) $x = \frac{2}{3}$

(B) $x = \frac{e^4}{6}$

(D) $x = \frac{e^4 – 2}{3}$

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
Apply the product property of logarithms: $\ln(a) + \ln(b) = \ln(ab)$.
Combine the terms: $\ln(3x \cdot 2) = 4$, which simplifies to $\ln(6x) = 4$.
Convert the natural logarithm to exponential form: $e^4 = 6x$.
Isolate $x$ by dividing both sides by 6: $x = \frac{e^4}{6}$.
Verify the domain: since $x > 0$, the solution is valid.

Question

What are all values of $x$ for which $\log_{7}(x-1) + 2 = \log_{7}(4x+6)$?

(A) $x = -\frac{5}{3}$

(B) $x = -4$

(D) there are no such values of $x$

▶️ Answer/Explanation

Detailed solution

Express the constant as a logarithm: $2 = \log_{7}(7^2) = \log_{7}(49)$.
Apply the product rule: $\log_{7}(x-1) + \log_{7}(49) = \log_{7}(49(x-1))$.
Set the arguments equal: $49(x-1) = 4x + 6$.
Expand and simplify: $49x – 49 = 4x + 6$.
Solve for $x$: $45x = 55$, which reduces to $x = \frac{11}{9}$.
Check domain constraints: $x-1 > 0 \implies x > 1$ and $4x+6 > 0 \implies x > -1.5$.
Since $\frac{11}{9} \approx 1.22$ satisfies $x > 1$, the solution is valid.
The correct option is (C).

Question

What are all values of $x$ for which $4^{2x+6} = 2^{3x}$?

(A) $x = -12$
(B) $x = -6$
(C) $x = \frac{3}{2}$
(D) $x = 6$

▶️ Answer/Explanation

Detailed solution

The correct answer is (A).
Rewrite the base $4$ as $2^2$ to get $(2^2)^{2x+6} = 2^{3x}$.
Apply the power of a power rule: $2^{2(2x+6)} = 2^{3x}$.
Since the bases are equal, set the exponents equal: $2(2x+6) = 3x$.
Distribute the $2$: $4x + 12 = 3x$.
Subtract $3x$ from both sides: $x + 12 = 0$.
Solve for $x$: $x = -12$.

Question

Consider the functions $h$ and $k$ given by $h(x) = 9^{(3x-1)}$ and $k(x) = 81^{(x+3)}$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $h$ and $k$?

(A) $-\frac{14}{3}$

(B) $\frac{6}{13}$

(D) $7$

▶️ Answer/Explanation

Detailed solution

The point of intersection occurs where $h(x) = k(x)$.
Set the equations equal: $9^{(3x-1)} = 81^{(x+3)}$.
Express $81$ as a power of $9$, specifically $81 = 9^2$.
Substitute to get $9^{(3x-1)} = (9^2)^{(x+3)}$.
Apply the power rule $(a^m)^n = a^{mn}$ to get $9^{(3x-1)} = 9^{2(x+3)}$.
Equate the exponents: $3x – 1 = 2(x + 3)$.
Simplify and solve: $3x – 1 = 2x + 6$.
Subtract $2x$ and add $1$ to both sides to find $x = 7$.
The correct option is (D).

Question

Consider the functions $m$ and $p$ given by $m(x) = 4^{(x^2 – 2x)}$ and $p(x) = \left( \frac{1}{16} \right)^{x-2}$. In the $xy$-plane, what are all $x$-coordinates of the points of intersection of the graphs of $m$ and $p$?

(A) $2$ only
(B) $2$ and $-2$
(C) $2$ and $-\frac{1}{2}$
(D) $2$ and $1$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set $m(x) = p(x)$, giving $4^{(x^2 – 2x)} = \left( \frac{1}{16} \right)^{x-2}$.
Express both sides with base $4$: $4^{(x^2 – 2x)} = (4^{-2})^{x-2}$.
Simplify the right side power: $4^{(x^2 – 2x)} = 4^{-2x + 4}$.
Equate the exponents: $x^2 – 2x = -2x + 4$.
Add $2x$ to both sides to simplify: $x^2 = 4$.
Solve for $x$: $x = 2$ and $x = -2$.
The correct option is (B).

Question

Consider the functions $h$ and $k$ given by $h(x) = 3^x$ and $k(x) = 9^{x-1}$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $h$ and $k$?

(A) $-1$

(B) $0$

(D) $2$

▶️ Answer/Explanation

The correct answer is (D) $2$.

Detailed solution:

To find the intersection, set the functions equal: $h(x) = k(x)$.
Substitute the expressions: $3^x = 9^{x-1}$.
Rewrite $9$ as a power of $3$: $3^x = (3^2)^{x-1}$.
Apply the power of a power rule: $3^x = 3^{2(x-1)}$.
Equate the exponents since the bases are the same: $x = 2(x-1)$.
Distribute the $2$: $x = 2x – 2$.
Subtract $x$ and add $2$ to solve: $x = 2$.

Question

Consider the functions $g$ and $h$ given by $g(x) = 4^x$ and $h(x) = \left(\frac{1}{2}\right)^{x+3}$. In the $xy$-plane, what is the $x$-coordinate of the point of intersection of the graphs of $g$ and $h$?

(A) $-6$

(B) $-2$

(D) $3$

▶️ Answer/Explanation

Detailed solution

To find the intersection, set the functions equal: $g(x) = h(x)$.
This gives the equation: $4^x = \left(\frac{1}{2}\right)^{x+3}$.
Express both sides with base $2$: $(2^2)^x = (2^{-1})^{x+3}$.
Simplify the exponents: $2^{2x} = 2^{-(x+3)}$.
Set the exponents equal to each other: $2x = -x – 3$.
Add $x$ to both sides: $3x = -3$.
Divide by $3$: $x = -1$.
The correct option is (C).

Question

The function $f$ is given by $f(x) = 3 + 2e^{-x}$. For which of the following values of $x$ is $f(x) = 11$?

(A) $x = 1 + \ln 4$

(B) $x = \frac{-\ln 8}{2}$

(D) $x = \frac{3 – \ln 11}{2}$

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $3 + 2e^{-x} = 11$.
Subtract $3$ from both sides to isolate the exponential term: $2e^{-x} = 8$.
Divide both sides by $2$ to get $e^{-x} = 4$.
Take the natural logarithm ($\ln$) of both sides: $-x = \ln 4$.
Multiply by $-1$ to solve for $x$: $x = -\ln 4$.
The correct option is (C).

Question

The function $g$ is given by $g(x) = 3^{x+1}$. For which of the following values of $x$ is $g(x) = 27$?

(A) $x = -\frac{2}{3}$
(B) $x = 2$
(C) $x = 3$
(D) $x = 4$

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
Given the function $g(x) = 3^{x+1}$, we set the equation to $3^{x+1} = 27$.
Express $27$ as a power of $3$, which is $3^3$.
Rewrite the equation as $3^{x+1} = 3^3$.
Since the bases are equal, set the exponents equal: $x + 1 = 3$.
Subtract $1$ from both sides to solve for $x$.
The result is $x = 2$.

Question

The function $h$ is given by $h(x) = 2^{3x-1}$. For which of the following values of $x$ is $h(x) = \frac{1}{16}$?

(A) $x = -4$

(B) $x = -1$

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

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $2^{3x-1} = \frac{1}{16}$.
Express $\frac{1}{16}$ as a power of 2: $\frac{1}{16} = \frac{1}{2^4} = 2^{-4}$.
Equate the exponents since the bases are the same: $3x – 1 = -4$.
Add 1 to both sides of the equation: $3x = -3$.
Divide both sides by 3 to solve for $x$: $x = -1$.
Therefore, the correct value is $x = -1$, which corresponds to option (B).

Question

The function $k$ is given by $k(x) = 3 \cdot 8^x$. For which of the following values of $x$ is $k(x) = 12$?

(A) $x = -\frac{2}{3}$

(B) $x = 0$

(D) $x = \frac{2}{3}$

▶️ Answer/Explanation

Detailed solution

Set the function equal to the target value: $3 \cdot 8^x = 12$.
Divide both sides by $3$ to isolate the exponential term: $8^x = 4$.
Express both $8$ and $4$ as powers of the common base $2$: $(2^3)^x = 2^2$.
Apply the power of a power rule: $2^{3x} = 2^2$.
Equate the exponents since the bases are the same: $3x = 2$.
Solve for $x$ by dividing by $3$: $x = \frac{2}{3}$.
The correct option is (D).

Question

(A) $\frac{1}{5}$

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

(D) $2$

▶️ Answer/Explanation

Detailed solution

Express $g(x)$ with base $2$: $g(x) = (2^2)^{x+1} = 2^{2x+2}$.
Set up the equation for $f(x) = 1$: $\frac{2^{2x+2}}{2^{3x}} = 1$.
Apply the quotient rule for exponents: $2^{(2x+2) – 3x} = 2^0$.
Simplify the exponent: $2^{-x+2} = 2^0$.
Equate the exponents: $-x + 2 = 0$.
Solve for $x$: $x = 2$.
The correct option is (D).

Question

Consider the functions $f$ and $g$ given by $f(x) = 9^{x-1}$ and $g(x) = \log_{3} x$. Let $h$ be the function given by $h(x) = g(f(x))$. In the $xy$-plane, on which intervals is $h(x) > 3$?

(A) $\left( \frac{3}{2}, \infty \right)$

(B) $\left( \frac{5}{2}, \infty \right)$ only

(D) $(27, \infty)$ only

▶️ Answer/Explanation

Detailed solution

The correct option is (B).
Substitute $f(x)$ into $g(x)$ to find $h(x) = \log_{3}(9^{x-1})$.
Rewrite $9$ as $3^2$, giving $h(x) = \log_{3}((3^2)^{x-1}) = \log_{3}(3^{2x-2})$.
Simplify using log properties to get $h(x) = 2x – 2$.
Set up the inequality $2x – 2 > 3$.
Add $2$ to both sides to get $2x > 5$.
Divide by $2$ to find $x > \frac{5}{2}$, which is the interval $\left( \frac{5}{2}, \infty \right)$.

Question (Calc Allowed)

The growth of bacteria in a culture is modeled by $y = 2.3e^{1.5t}$, where $t$ is measured in days. At what time $t$ is the number of bacteria approximately $210$?

(A) $t = 3.009 \text{ days}$

(B) $t = 3.063 \text{ days}$

(D) $t = 5.758 \text{ days}$

▶️ Answer/Explanation

Detailed solution

Set the bacteria count $y = 210$ in the given model: $210 = 2.3e^{1.5t}$.
Divide both sides by $2.3$ to isolate the exponential term: $\frac{210}{2.3} = e^{1.5t}$.
Calculate the ratio: $e^{1.5t} \approx 91.304$.
Take the natural logarithm ($\ln$) of both sides: $\ln(91.304) = 1.5t$.
Solve for $t$ by dividing by $1.5$: $t = \frac{\ln(91.304)}{1.5}$.
Using a calculator: $t \approx \frac{4.5142}{1.5} \approx 3.0094$.
The time $t$ is approximately $3.009 \text{ days}$.
The correct option is (A).