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.
▶️ Answer/Explanation
\( g(x) = 2 \) ⇒ \( \ln(f(x)) = 2 \) ⇒ \( f(x) = e^2 \) by definition of natural log.
✅ Answer: (B)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
\( g(x) < 0 \) ⇒ \( \ln\frac{3x+1}{x^2 + x – 2} < 0 \) ⇒ \( \frac{3x+1}{x^2 + x – 2} < 1 \), 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)
▶️ Answer/Explanation
\( d(N) > 140 \) ⇒ \( 10 \log_{10} \left( \frac{N}{10^{-12}} \right) > 140 \)
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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.
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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:
\( x = \frac{e^4}{6} \)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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)**.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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: $x = \pm 4\sqrt{2}$
▶️ Answer/Explanation
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.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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 $x$: $x = -12$.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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: $x = 2$.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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 $x = 2$.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
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 $\left( \frac{5}{2}, \infty \right)$.
▶️ Answer/Explanation
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).
▶️ Answer/Explanation
(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 \( k \in \mathbb{Z} \)
▶️ Answer/Explanation
(A)(i) Find \( h(3) \)
The function is defined as \( h(x) = g(f(x)) \).
First, we find the value of the inner function, \( f(3) \).
According to the problem text and graph, the point \( (3, 2) \) lies on the graph of \( f \). Therefore, \( f(3) = 2 \).
Now, substitute this value into the outer function \( g(x) \):
\( h(3) = g(2) \)
Using the definition \( g(x) = 2 + 3\ln x \):
\( g(2) = 2 + 3\ln(2) \)
Using a calculator to approximate \( \ln(2) \approx 0.693 \):
\( g(2) = 2 + 3(0.6931…) \approx 2 + 2.079 = 4.079 \)
Answer: \( h(3) \approx 4.079 \)
(A)(ii) Find real zeros of \( f \)
A real zero of a function occurs where the graph intersects the x-axis (where \( f(x) = 0 \)).
Looking at the provided graph, the curve intersects the x-axis at \( x = -1 \).
The problem text confirms the point \( (-1, 0) \) is on the graph.
There are no other intersections with the x-axis shown.
Answer: \( x = -1 \)
(B)(i) Find \( x \) for \( g(x) = e \)
Set up the equation:
\( 2 + 3\ln x = e \)
Subtract 2 from both sides:
\( 3\ln x = e – 2 \)
Divide by 3:
\( \ln x = \frac{e – 2}{3} \)
Convert from logarithmic to exponential form to solve for \( x \):
\( x = e^{\left(\frac{e – 2}{3}\right)} \)
Approximating the value (using \( e \approx 2.718 \)):
\( x \approx e^{0.2394} \)
Answer: \( x \approx 1.271 \)
(B)(ii) End behavior of \( g \)
We need to evaluate the limit as \( x \to \infty \) for \( g(x) = 2 + 3\ln x \).
As \( x \) increases without bound (\( x \to \infty \)), the natural logarithm function \( \ln x \) also increases without bound (\( \ln x \to \infty \)).
Multiplying by 3 and adding 2 does not change the unbounded nature.
Answer: \( \lim_{x \to \infty} g(x) = \infty \)
(C)(i) Is \( f \) invertible?
Answer: Yes, \( f \) is invertible.
(C)(ii) Reason
A function is invertible if and only if it is one-to-one. This can be verified visually using the Horizontal Line Test.
Looking at the graph of \( f(x) \):
1. The function is strictly decreasing on both branches of its domain (\( x < 1 \) and \( x > 1 \)).
2. The range of the left branch appears to be \( (-\infty, 1) \) and the range of the right branch appears to be \( (1, \infty) \).
Because the y-values do not repeat (no horizontal line intersects the graph more than once), the function is one-to-one.
Therefore, \( f \) has an inverse.
▶️ Answer/Explanation
(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 \( -6.5 \)