$\frac{\log_25\cdot\log_65 + \log_35\cdot\log_65}{\log_25\cdot\log_35}$

Question

I see that every elements has 5 on them.

So, i make $\log_25$ into $\frac{1}{\log_52}$,

$\frac{1}{\log_52\cdot\log_56\cdot\log_56} + \frac{1}{\log_53\cdot\log_56\cdot\log_56}$

I simplify it i get $({\log_65})^2$ Is it right ? But i cannot evaluate the value of it, no calculator allowed.

Answer 1

\begin{align} \frac{\log_25\cdot\log_65+\log_35\cdot\log_65}{\log_25\cdot\log_35}&=\frac{\log_25\cdot\log_65}{\log_25\cdot\log_35}+\frac{\log_35\cdot\log_65}{\log_25\cdot\log_35}\\ &=\frac{\log_65}{\log_35}+\frac{\log_65}{\log_25}\\ &=\frac{\log_53}{\log_56}+\frac{\log_52}{\log_56}\\ &=\frac{\log_53+\log_52}{\log_56}\\ &=\frac{\log_5(3\cdot 2)}{\log_56}\\ &=\frac{\log_56}{\log_56}\\ &=1 \end{align}

Answer 2

\begin{align} \frac{\log_25\cdot\log_65+\log_35\cdot\log_65}{\log_25\cdot\log_35}&=\frac{\log_65\cdot(\log_25+\log_35)}{\log_25\cdot\log_35}\\ &=\frac{\log_65\cdot(\frac{\log_53+\log_52}{\log_52\cdot\log_53})}{\log_25\cdot\log_35}\\ &=\log_65\cdot(\frac{\log_53+\log_52}{\log_52\cdot\log_53})\cdot\log_52\cdot\log_53\\ &=\log_65\cdot\log_56\\ &=1 \end{align}