AP Precalculus -1.11 Equivalent Forms of Polynomial and Rational Expressions- FRQ Exam Style Questions - Effective Fall 2023

(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 \( -6.5 \)