[Algebra]
Problem: Consider the system of equations

$$
\begin{aligned}
& \log _{4} x+\log _{8}(y z)=2 \\
& \log _{4} y+\log _{8}(x z)=4 \\
& \log _{4} z+\log _{8}(x y)=5 .
\end{aligned}
$$

Given that $x y z$ can be expressed in the form $2^{k}$, compute $k$.

Solution: Note that for $n>0, \log _{4} n=\log _{64} n^{3}$ and $\log _{8} n=\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields

$$
\begin{aligned}
& \log _{4}(x y z)+\log _{8}\left(x^{2} y^{2} z^{2}\right)=11 \\
\Longrightarrow & \log _{64}(x y z)^{3}+\log _{64}(x y z)^{4}=11 \\
\Longrightarrow & \log _{64}(x y z)^{7}=11 \\
\Longrightarrow & 7 \log _{64}(x y z)=11 .
\end{aligned}
$$

The last equation is equivalent to $x y z=64^{11 / 7}=2^{66 / 7}$, hence the desired value of $k$ is $\frac{\mathbf{6 6}}{\mathbf{7}}$.

Final answer: $\frac{66}{7}$