### Instruction: Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x).$.### Response: This problem looks interesting because it involves logs of logs and different bases.
I wonder if there is a way to simplify it using some properties of logarithms.
One thing I notice is that $\log_8 x = \frac{\log_2 x}{\log_2 8}$, since changing the base of a logarithm is equivalent to dividing by the new base's logarithm.
Maybe I can use this to rewrite the given equation in terms of $\log_2 x$ only.
If I substitute $\log_8 x = \frac{\log_2 x}{\log_2 8}$ in both sides, I get $\log_2 (\frac{\log_2 x}{\log_2 8}) = \frac{\log_2 (\log_2 x)}{\log_2 8}$.
Now I can simplify this by using another property of logarithms, namely that $\log_a (b/c) = \log_a b - \log_a c$.
Applying this to both sides, I get $\log_2 (\log_2 x) - \log_2 (\log_2 8) = \frac{\log_2 (\log_2 x)}{\log_2 8} - \frac{\log_2 (\log_2 8)}{\log_2 8}$.
This looks more manageable, and I can simplify it further by noting that $\log_2 (\log_2 8) = \log_2 3$, since $8 = 2^3$.
Also, I can multiply both sides by $\log_2 8$ to clear the fractions.
Doing so, I get $\log_2 8 \log_2 (\log_2 x) - \log_2 8 \log_2 3 = \log_2 (\log_2 x) - \log_2 3$.
Now I can rearrange the terms to get all the $\log_2 (\log_2 x)$ terms on one side and all the constants on the other.
This gives me $\log_2 8 \log_2 (\log_2 x) - \log_2 (\log_2 x) = \log_2 8 \log_2 3 - \log_2 3$.
I can factor out $\log_2 (\log_2 x)$ from the left side and $\log_2 3$ from the right side.
This gives me $\log_2 (\log_2 x) (\log_2 8 - 1) = \log_2 3 (\log_2 8 - 1)$.
I notice that both sides have the same factor of $\log_2 8 - 1$, which means I can divide both sides by it, as long as it is not zero.
I check that $\log_2 8 - 1$ is not zero, since $\log_2 8 = 3$ and $3 - 1 = 2$, which is nonzero.
So I can divide both sides by $\log_2 8 - 1$ and get $\log_2 (\log_2 x) = \log_2 3$.
This means that $\log_2 x$ must be equal to $2^{\log_2 3}$, since raising both sides to the base 2 is the inverse operation of taking the logarithm base 2.
But $2^{\log_2 3}$ is just 3, by the definition of logarithm.
So I have $\log_2 x = 3$, and I can raise both sides to the base 2 again to get $x = 2^3 = 8$.
This is the value of x that satisfies the given equation.
Now I can plug it into the expression $(\log_2 x)^2$ and evaluate it.
This gives me $(\log_2 8)^2 = 3^2 = 9$.