Question

determine the rule for $g \circ f(x)$ given the functions below.
$f(x) = -\log_{2} x$
$g(x) = 16^{x}$

Explanation:

Step1: Recall the composition of functions

The composition \( g \circ f(x) \) means \( g(f(x)) \). So we need to substitute \( f(x) \) into \( g(x) \).

Step2: Substitute \( f(x) \) into \( g(x) \)

Given \( f(x)=-\log_{2}x \) and \( g(x) = 16^{x}\), then \( g(f(x))=16^{-\log_{2}x} \).

Step3: Simplify the exponent using logarithm and exponent properties

We know that \( 16 = 2^{4} \), so \( 16^{-\log_{2}x}=(2^{4})^{-\log_{2}x} \).
Using the power of a power rule \( (a^{m})^{n}=a^{mn} \), we get \( (2^{4})^{-\log_{2}x}=2^{-4\log_{2}x} \).
Also, recall the property \( a^{\log_{a}b}=b \) and \( n\log_{a}b=\log_{a}b^{n} \). So \( 2^{-4\log_{2}x}=2^{\log_{2}x^{-4}} \).
By the property \( a^{\log_{a}b}=b \), \( 2^{\log_{2}x^{-4}}=x^{-4} \).
And \( x^{-4}=\frac{1}{x^{4}} \).

Answer:

\( g\circ f(x)=\frac{1}{x^{4}} \)