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

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

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

AP Precalculus – FRQ Exam Style Questions- All Topics

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- FRQ Exam Style Questions - Effective Fall 2023