2730.33 – Love Those Logarithms

Let $x = \log_{\sqrt[6]{2}}{\left( \sqrt[3]{32} \right)}$. What this means is that

$$\left( \sqrt[6]{2} \right)^x = \sqrt[3]{32}.$$

By manipulation of both sides of this equality, one can discover $x$ in a simpler form. Thus,

$$\begin{aligned}\left( \sqrt[6]{2} \right)^x &= \sqrt[3]{32} \\2^{\left( \dfrac{1}{6} \right) ^ x} &= \sqrt[3]{2^5} \\2^{ \left( \dfrac{x}{6} \right)} &= 2^{\left(\dfrac{5}{3}\right)}\end{aligned}$$

Therefore $\dfrac{x}{6} = \dfrac{5}{3}$ from which follows $x = 10$.

Were we expecting such a simple answer for such a messy-looking problem?