[Algebra]
Problem: Determine all values of $x$ such that $\log _{2 x}(48 \sqrt[3]{3})=\log _{3 x}(162 \sqrt[3]{2})$.

Solution: Since the base of a logarithm must be positive and cannot equal 1 , then $x>0$ and $x \neq \frac{1}{2}$ and $x \neq \frac{1}{3}$.

This tells us that $\log 2 x$ and $\log 3 x$ exist and do not equal 0 , which we will need shortly when we apply the change of base formula.

We note further that $48=2^{4} \cdot 3$ and $162=3^{4} \cdot 2$ and $\sqrt[3]{3}=3^{1 / 3}$ and $\sqrt[3]{2}=2^{1 / 3}$. Using logarithm rules, the following equations are equivalent:

$$
\begin{aligned}
\log _{2 x}(48 \sqrt[3]{3}) & =\log _{3 x}(162 \sqrt[3]{2}) \\
\frac{\log \left(2^{4} \cdot 3 \cdot 3^{1 / 3}\right)}{\log 2 x} & =\frac{\log \left(3^{4} \cdot 2 \cdot 2^{1 / 3}\right)}{\log 3 x} \quad \text { (change of base formula) } \\
\frac{\log \left(2^{4} \cdot 3^{4 / 3}\right)}{\log 2+\log x} & =\frac{\log \left(3^{4} \cdot 2^{4 / 3}\right)}{\log 3+\log x} \quad(\log a b=\log a+\log b) \\
\frac{\log \left(2^{4}\right)+\log \left(3^{4 / 3}\right)}{\log 2+\log x} & =\frac{\log \left(3^{4}\right)+\log \left(2^{4 / 3}\right)}{\log 3+\log x} \quad(\log a b=\log a+\log b) \\
\frac{4 \log 2+\frac{4}{3} \log 3}{\log 2+\log x} & =\frac{4 \log 3+\frac{4}{3} \log 2}{\log 3+\log x} \quad\left(\log \left(a^{c}\right)=c \log a\right)
\end{aligned}
$$

Cross-multiplying, we obtain

$$
\left(4 \log 2+\frac{4}{3} \log 3\right)(\log 3+\log x)=\left(4 \log 3+\frac{4}{3} \log 2\right)(\log 2+\log x)
$$

Expanding the left side, we obtain

$$
4 \log 2 \log 3+\frac{4}{3}(\log 3)^{2}+\left(4 \log 2+\frac{4}{3} \log 3\right) \log x
$$

Expanding the right side, we obtain

$$
4 \log 3 \log 2+\frac{4}{3}(\log 2)^{2}+\left(4 \log 3+\frac{4}{3} \log 2\right) \log x
$$

Simplifying and factoring, we obtain the following equivalent equations:

$$
\begin{aligned}
\frac{4}{3}(\log 3)^{2}-\frac{4}{3}(\log 2)^{2} & =\log x\left(4 \log 3+\frac{4}{3} \log 2-4 \log 2-\frac{4}{3} \log 3\right) \\
\frac{4}{3}(\log 3)^{2}-\frac{4}{3}(\log 2)^{2} & =\log x\left(\frac{8}{3} \log 3-\frac{8}{3} \log 2\right) \\
(\log 3)^{2}-(\log 2)^{2} & =2 \log x(\log 3-\log 2) \\
\log x & =\frac{(\log 3)^{2}-(\log 2)^{2}}{2(\log 3-\log 2)} \\
\log x & =\frac{(\log 3-\log 2)(\log 3+\log 2)}{2(\log 3-\log 2)} \\
\log x & =\frac{\log 3+\log 2}{2} \\
\log x & =\frac{1}{2} \log 6 \\
\log x & =\log (\sqrt{6})
\end{aligned}
$$

and so $x=\sqrt{6}$.

Final answer: $\sqrt{6}$