2730.33 – Love Those Logarithms

Find $\log_{\sqrt[6]{2}}{\left( \sqrt[3]{32} \right)}$ . Note the unusual base of this logarithm.

Solution

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 = 2^{\left( \frac{1}{6} \right) ^ x} = 2^{ \left( \frac{x}{6} \right)} &=& \sqrt[3]{32}, \\ &=& \sqrt[3]{2^5} \\ &=& 2^{\frac{5}{3}}. \end{aligned}

In summary,

$2^{ \left( \frac{x}{6} \right)} = 2^{\frac{5}{3}}.$

Therefore $x/6 = 5/3$ from which follows $x = 10$ .

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