<s>[INST] Expansion of Logarithms with Cube Roots Does the following expand to the following
$$ \log_6(11^6\sqrt[3]{12}) $$
= $ 6\log_6(11) + \log_6 (\sqrt[3]{12})$ 
 [/INST] First note that
\[ \sqrt[n]{x}=x^{\frac{1}{n}} \]
\[ \log_{b}(xy)=\log_{b}(x)+\log_{b}(y) \]
\[ \log_{b}(x^{n})=n\log_{b}(x) \]
\[ \log_{b}(x)=\frac{\log_{10}(x)}{\log_{10}(b)}=\frac{\log(x)}{\log(b)} \]
So then
\[
\log_{6}(11^{6}\cdot\sqrt[3]{12})=\log_{6}(11^{6}\cdot 12^{\frac{1}{3}})=\log_{6}(11^{6})+\log_{6}(12^{\frac{1}{3}})=6\log_{6}(11)+\frac{1}{3}\log_{6}(12)
\]
Or perhaps
\[
\frac{6\log(11)}{\log(6)}+\frac{\log(12)}{3\log(6)}
\]
</s>