Expansion of Logarithms with Cube Roots

Question

Does the following expand to the following $$ \log_6(11^6\sqrt[3]{12}) $$

= $ 6\log_6(11) + \log_6 (\sqrt[3]{12})$

Answer 1

Yes and this is equal to $$6\log_6(11)+\frac{1}{3}(\log(2)+1)$$

Answer 2

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 \[ \frac{6\log(11)}{\log(6)}+\frac{\log(12)}{3\log(6)} \]