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Question 1209955: Suppose that f(x) and g(x) are functions which satisfy
f(g(x)) = x^2 \quad \text{and} \quad g(f(x)) = x^4
for all x \ge 1. If g(16) = 1, then compute \log_2 g(2).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $f(x)$ and $g(x)$ be functions such that $f(g(x)) = x^2$ and $g(f(x)) = x^4$ for all $x \ge 1$. We are given that $g(16) = 1$. We want to find $\log_2 g(2)$.
We have $f(g(x)) = x^2$ and $g(f(x)) = x^4$.
Let's consider $g(f(g(x)))$. Using the first equation, $f(g(x)) = x^2$, so
$$g(f(g(x))) = g(x^2).$$
Using the second equation, $g(f(x)) = x^4$, so
$$g(f(g(x))) = (g(x))^4.$$
Therefore, we have $g(x^2) = (g(x))^4$.
We are given that $g(16) = 1$. Let's use the equation $g(x^2) = (g(x))^4$ with $x = 4$:
$$g(4^2) = g(16) = (g(4))^4.$$
Since $g(16) = 1$, we have $(g(4))^4 = 1$. Since $g(x)$ is a function and $g(4)$ must be positive, we have $g(4) = 1$.
Now, let's use the equation $g(x^2) = (g(x))^4$ with $x = 2$:
$$g(2^2) = g(4) = (g(2))^4.$$
Since $g(4) = 1$, we have $(g(2))^4 = 1$. Since $g(2)$ must be positive, we have $g(2) = 1$.
Finally, we want to compute $\log_2 g(2)$. Since $g(2) = 1$, we have
$$\log_2 g(2) = \log_2 1 = 0.$$
Final Answer: The final answer is $\boxed{0}$.
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