Answer:- The functions f(x) and g(x) are equivalent.
Explanation:-
Given functions:- [tex]f(x)=\sqrt{16^x}[/tex] and [tex]g(x)=\sqrt[3]{64^x }[/tex]
Simplify the functions by using law of exponents
[tex](a^n)^m=a^{nm}[/tex]
Thus [tex]f(x)=\sqrt{16^x}=\sqrt{(4^2)^x}=\sqrt{4^{2x}} =\sqrt{(4^x)^2}=4^x[/tex]
and [tex]g(x)=\sqrt[3]{64^x}=\sqrt[3]{(4^3)^x}= \sqrt[3]{4^{3x}}= \sqrt[3]{(4^x)^3}=4^x[/tex]
⇒f(x)=g(x)
Therefore ,the functions f(x) and g(x) are equivalent.
It is clear that g(x) equal to f(x).
A quantity representing the exponent to which a given number or letter is to be raised is usually expressed as a raised symbol beside the number or letter. It is expressed as [tex]\rm a^{x}[/tex] is a variable and a is constant.
Given
[tex]\rm f(x) = \sqrt{16^{x}}\ and\ g(x) = \sqrt[3]{64^{x}}[/tex]two exponents.
To find
The functions f(x) and g(x) are equivalent.
[tex]\rm f(x) = \sqrt{16^{x}}\ and\ g(x) = \sqrt[3]{64^{x}}[/tex]two exponents can be written as
[tex]\rm f(x) = \sqrt{16^{x}}\ = 2^{\frac{4x}{2}} = 2^{2x}[/tex]
and the other function will be
[tex]\rm g(x) = \sqrt[3]{64^{x} } \ = 2^{\frac{6x}{3}} = 2^{2x}[/tex]
It is clear that g(x) equal to f(x) because g(x) is having power equal than f(x).
Thus, It is clear that g(x) is equal to f(x).
More about the exponent link is given below.
https://brainly.com/question/5497425
