[Algebra]
Problem: Determine all real values of $x$ for which

$$
\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}=4
$$

Solution: Let $f(x)=\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}$.

Using logarithm laws,

$$
\begin{aligned}
\log _{2} x \cdot \log _{2}(4 x)+1 & =\log _{2} x\left(\log _{2} 4+\log _{2} x\right)+1 \\
& =\log _{2} x\left(2+\log _{2} x\right)+1 \quad\left(\text { since } 2^{2}=4\right) \\
& =\left(\log _{2} x\right)^{2}+2 \cdot \log _{2} x+1 \\
& =\left(\log _{2} x+1\right)^{2}
\end{aligned}
$$

and

$$
\begin{aligned}
\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9 & =\log _{2} x\left(\log _{2} x-\log _{2} 64\right)+9 \\
& =\log _{2} x\left(\log _{2} x-6\right)+9 \quad\left(\text { since } 2^{6}=64\right) \\
& =\left(\log _{2} x\right)^{2}-6 \log _{2} x+9 \\
& =\left(\log _{2} x-3\right)^{2}
\end{aligned}
$$

Therefore,

$f(x)=\sqrt{\log _{2} x \cdot \log _{2}(4 x)+1}+\sqrt{\log _{2} x \cdot \log _{2}\left(\frac{x}{64}\right)+9}=\sqrt{\left(\log _{2} x+1\right)^{2}}+\sqrt{\left(\log _{2} x-3\right)^{2}}$

Before proceeding, we recall that if $a \leq 0$, then $\sqrt{a^{2}}=-a$ and if $a>0$, then $\sqrt{a^{2}}=a$.

When $\log _{2} x \leq-1$, we know that $\log _{2} x+1 \leq 0$ and $\log _{2} x-3<0$, and so

$$
f(x)=-\left(\log _{2} x+1\right)-\left(\log _{2} x-3\right)=2-2 \log _{2} x
$$

When $-1<\log _{2} x \leq 3$, we know that $\log _{2} x+1>0$ and $\log _{2} x-3 \leq 0$, and so

$$
f(x)=\left(\log _{2} x+1\right)-\left(\log _{2} x-3\right)=4
$$

When $\log _{2} x>3$, we know that $\log _{2} x+1 \geq 0$ and $\log _{2} x-3>0$, and so

$$
f(x)=\left(\log _{2} x+1\right)+\left(\log _{2} x-3\right)=2 \log _{2} x-2
$$

We want to find all values of $x$ for which $f(x)=4$.

When $\log _{2} x \leq-1, f(x)=2-2 \log _{2} x=4$ exactly when $\log _{2} x=-1$.

When $-1<\log _{2} x \leq 3, f(x)$ is always equal to 4 .

When $\log _{2} x>3, f(x)=2 \log _{2} x-2=4$ exactly when $\log _{2} x=3$.

Therefore, $f(x)=4$ exactly when $-1 \leq \log _{2} x \leq 3$, which is true exactly when $\frac{1}{2} \leq x \leq 8$. (It seems surprising that the solution to this equation is actually an interval of values, rather than a finite number of specific values.)

Final answer: $[\frac{1}{2}, 8]$