What are all real numbers $x>0$ for which $\log _{2}\left(x^{2}\right)+2 \log _{x} 8=\frac{392}{\log _{2}\left(x^{3}\right)+20 \log _{x}(32)}$ ?
Solution 1

Since $\log _{a}\left(b^{c}\right)=c \log _{a} b$, then

$$
\begin{aligned}
\log _{2}\left(x^{2}\right) & =2 \log _{2} x \\
\log _{2}\left(x^{3}\right) & =3 \log _{2} x \\
2 \log _{x} 8 & =2 \log _{x}\left(2^{3}\right)=2 \cdot 3 \log _{x} 2=6 \log _{x} 2 \\
20 \log _{x}(32) & =20 \log _{x}\left(2^{5}\right)=20 \cdot 5 \log _{x} 2=100 \log _{x} 2
\end{aligned}
$$

Let $t=\log _{2} x$.

Since $x$ is also the base of a logarithm in the equation, then $x \neq 1$, which means that $t \neq 0$. Therefore,

$$
\log _{x} 2=\frac{\log 2}{\log x}=\frac{1}{(\log x) /(\log 2)}=\frac{1}{\log _{2} x}=\frac{1}{t}
$$

Starting with the original equation, we obtain the following equivalent equations:

$$
\begin{aligned}
\log _{2}\left(x^{2}\right)+2 \log _{x} 8 & =\frac{392}{\log _{2}\left(x^{3}\right)+20 \log _{x}(32)} \\
2 \log _{2} x+6 \log _{x} 2 & =\frac{392}{3 \log _{2} x+100 \log _{x} 2} \\
2 t+\frac{6}{t} & =\frac{392}{3 t+\frac{100}{t}} \quad\left(\text { using } t=\log _{2} x\right) \\
\frac{2 t^{2}+6}{t} & =\frac{392 t}{3 t^{2}+100} \quad(\text { since } t \neq 0) \\
\frac{t^{2}+3}{t} & =\frac{196 t}{3 t^{2}+100} \\
\left(t^{2}+3\right)\left(3 t^{2}+100\right) & =196 t^{2} \\
3 t^{4}+109 t^{2}+300 & =196 t^{2} \\
t^{4}-87 t^{2}+300 & =0 \\
\left(t^{2}-25\right)\left(t^{2}-4\right) & =0 \\
(t+5)(t-5)(t+2)(t-2) & =0
\end{aligned}
$$

Therefore, $t=-5$ or $t=5$ or $t=-2$ or $t=2$.

Thus, $\log _{2} x=-5$ or $\log _{2} x=5$ or $\log _{2} x=-2$ or $\log _{2} x=2$.

Thus, $x=2^{-5}=\frac{1}{32}$ or $x=2^{5}=32$ or $x=2^{-2}=\frac{1}{4}$ or $x=2^{2}=4$.

We can check by substitution that each of these values of $x$ is indeed a solution to the original equation.

Solution 2

Let $a=\log _{2} x$ and $b=\log _{x} 2$.

Then $2^{a}=x$ and $x^{b}=2$, which gives $2^{a b}=\left(2^{a}\right)^{b}=x^{b}=2$.

Since $2^{a b}=2$, then $a b=1$.

As in Solution 1,

$$
\begin{aligned}
\log _{2}\left(x^{2}\right) & =2 \log _{2} x=2 a \\
\log _{2}\left(x^{3}\right) & =3 \log _{2} x=3 a \\
2 \log _{x} 8 & =6 \log _{x} 2=6 b \\
20 \log _{x}(32) & =100 \log _{x} 2=100 b
\end{aligned}
$$

Starting with the original equation, we thus obtain the following equivalent equations:

$$
\begin{aligned}
\log _{2}\left(x^{2}\right)+2 \log _{x} 8 & =\frac{392}{\log _{2}\left(x^{3}\right)+20 \log _{x}(32)} \\
2 a+6 b & =\frac{392}{3 a+100 b} \\
(2 a+6 b)(3 a+100 b) & =392 \\
(a+3 b)(3 a+100 b) & =196 \\
3 a^{2}+109 a b+300 b^{2} & =196 \\
3 a^{2}+60 a b+300 b^{2} & =196-49 a b \\
3 a^{2}+60 a b+300 b^{2} & =147 \quad \quad \text { (since } a b=1) \\
a^{2}+20 a b+100 b^{2} & =49 \quad \\
(a+10 b)^{2} & =7^{2}
\end{aligned}
$$

and so $a+10 b=7$ or $a+10 b=-7$.

Since $a b=1$, then $b=\frac{1}{a}$ and so $a+\frac{10}{a}=7$ or $a+\frac{10}{a}=-7$.

These equations give $a^{2}+10=7 a$ (or equivalently $a^{2}-7 a+10=0$ ) and $a^{2}+10=-7 a$ (or equivalently $\left.a^{2}+7 a+10=0\right)$.

Factoring, we obtain $(a-2)(a-5)=0$ or $(a+2)(a+5)=0$ and so $a=2,5,-2,-5$.

Since $a=\log _{2} x$, then $x=2^{-5}=\frac{1}{32}$ or $x=2^{5}=32$ or $x=2^{-2}=\frac{1}{4}$ or $x=2^{2}=4$.

We can check by substitution that each of these values of $x$ is indeed a solution to the original equation.

ANSWER: $x=\frac{1}{32}, 32, \frac{1}{4}, 4$