[Algebra]
Problem: Determine all real numbers $x>0$ for which

$$
\log _{4} x-\log _{x} 16=\frac{7}{6}-\log _{x} 8
$$

Solution: Note that $x \neq 1$ since 1 cannot be the base of a logarithm. This tells us that $\log x \neq 0$. Using the fact that $\log _{a} b=\frac{\log b}{\log a}$ and then using other logarithm laws, we obtain the following equivalent equations:

$$
\begin{aligned}
\log _{4} x-\log _{x} 16 & =\frac{7}{6}-\log _{x} 8 \\
\frac{\log x}{\log 4}-\frac{\log 16}{\log x} & =\frac{7}{6}-\frac{\log 8}{\log x} \quad(\text { note that } x \neq 1, \text { so } \log x \neq 0) \\
\frac{\log x}{\log 4} & =\frac{7}{6}+\frac{\log 16-\log 8}{\log x} \\
\frac{\log x}{\log \left(2^{2}\right)} & =\frac{7}{6}+\frac{\log \left(\frac{16}{8}\right)}{\log x} \\
\frac{\log x}{2 \log 2} & =\frac{7}{6}+\frac{\log 2}{\log x} \\
\frac{1}{2}\left(\frac{\log x}{\log 2}\right) & =\frac{7}{6}+\frac{\log 2}{\log x}
\end{aligned}
$$

Letting $t=\frac{\log x}{\log 2}=\log _{2} x$ and noting that $t \neq 0$ since $x \neq 1$, we obtain the following equations equivalent to the previous ones:

$$
\begin{aligned}
\frac{t}{2} & =\frac{7}{6}+\frac{1}{t} \\
3 t^{2} & =7 t+6 \quad(\text { multiplying both sides by } 6 t) \\
3 t^{2}-7 t-6 & =0 \\
(3 t+2)(t-3) & =0
\end{aligned}
$$

Therefore, the original equation is equivalent to $t=-\frac{2}{3}$ or $t=3$.

Converting back to the variable $x$, we obtain $\log _{2} x=-\frac{2}{3}$ or $\log _{2} x=3$, which gives $x=2^{-2 / 3}$ or $x=2^{3}=8$.

Final answer: $2^{-2 / 3}$, $8$