Question

Solve $\frac{loga+logbc}{loga(1+lo{g}_{a}bc)}$

Answer 1

$\frac{\log a+\log bc}{\log a(1+{\log }_{a}bc)}$

Changing base of ${\log }_{a}bc$

${\log }_{a}bc=\frac{\log bc}{\log a}$

Thus, $\frac{\log a+\log bc}{\log a(1+{\log }_{a}bc)}=\frac{\log a+\log bc}{\log a(1+\frac{\log bc}{\log a})}$

$\Rightarrow \frac{\log a+\log bc}{\log a\left(\frac{\log a+\log bc}{\log a}\right)}$

$\Rightarrow \frac{\log a+\log bc}{\left(\log a+\log bc\right)}$

$\Rightarrow 1$