Logarithm with nth root

Question

I made it but the result is very strange. I want every step to the result

$$ \large 6\log_{10}\frac{\sqrt2}{\sqrt[3]{3+\sqrt5}} $$

Answer 1

I assume that the question is how to expand the logarithm. Note that $$ 6 \log_{10}\left(\frac{\sqrt 2}{\sqrt[3]{3 + \sqrt{5}}} \right) =\\ 6 \log_{10}(\sqrt 2) - 6 \log_{10}(\sqrt[3]{3+\sqrt 5}) =\\ 6 \log_{10}(2^{1/2}) - 6 \log_{10}((3+\sqrt 5)^{1/3})=\\ 6 \cdot \frac 12 \log_{10}2 - 6 \cdot \frac 13\log_{10}(3+\sqrt 5) =\\ 3 \log_{10}2 - 2\log_{10}(3+\sqrt 5) $$

Answer 2

Here is an alternate approach from the first answer: $$ \begin{split} 6\log_{10}\frac{\sqrt2}{\sqrt[3]{3+\sqrt5}} &= \log_{10} \left(\frac{\sqrt2}{\sqrt[3]{3+\sqrt5}}\right)^6\\ &= \log_{10}\frac{2^3}{\left(3+\sqrt5\right)^2} \\ &= \log_{10}\frac{8}{14 + 6\sqrt5} \\ &= \log_{10}\frac{4}{7 + 3\sqrt5} \\ &= \log_{10}\frac{4}{7 + 3\sqrt5} \frac{}{7-3\sqrt5} \\ &= \log_{10}\frac{4(7-3\sqrt5)}{49 - 45} \\ &= \log_{10}(7-3\sqrt5) \end{split} $$