Suppose that $f (x)$ $f ( x )$ and $g (x)$ $g ( x )$ are functions which satisfy $f (g (x)) = x^{2}$ $f ( g ( x ) ) = x^2$ and $g (f (x)) = x^{3}$ $g ( f ( x ) ) = x^3$ for all $x \geq 1$ $x ≥ 1$ . If $g (16) = 16$ $g ( 16 ) = 16$ , then compute ${log}_{2} g (4)$ $log_2 g ( 4 )$ . (You may assume that $f (x) \geq 1$ $f ( x ) ≥ 1$ and $g (x) \geq 1$ $g ( x ) ≥ 1$ for all $x \geq 1$ $x ≥ 1$ .)?

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Answer 1 · 2018-02-14T10:49:22

$\frac{4}{3}$ $4/3$

Consider $g (f (g (4)))$ $g(f(g(4)))$

Since $f (g (4)) = 4^{2} = 16$ $f(g(4))=4^2=16$ , this is equal to $g (16) = 16$ $g(16)=16$ .

On the other hand, since $g (f (x)) = x^{3}$ $g(f(x))=x^3$ , we see that this is ${(g (4))}^{3}$ $(g(4))^3$ .

So

${(g (4))}^{3} = 16 \Rightarrow 3 {log}_{2} (g (4)) = {log}_{2} 16 = 4$ $(g(4))^3=16 implies 3 log_2(g(4)) = log_2 16 = 4$