Question

If $a={\log }_{24}12$, $b={\log }_{36}24$, $c={\log }_{48}36$ then show that $1+abc=2bc$.

Answer 1

$a={\log }_{24}\left(12\right)b={\log }_{36}\left(24\right)c={\log }_{48}\left(36\right)$

$1+abc=1+{\log }_{24}\left(12\right){\log }_{36}\left(24\right){\log }_{48}\left(36\right)$

$\Rightarrow 1+abc=1+\frac{\log 12}{\log 24}\times \frac{\log 24}{\log 36}\times \frac{\log 36}{\log 48}$

$\Rightarrow 1+abc=1+\frac{\log 12}{\log 48}=\frac{\log 48+\log 12}{\log 48}$

$\Rightarrow 1+abc=\frac{\log (48\times 12)}{\log 48}=\frac{\log (24{)}^2}{\log 48}=2\frac{\log 24}{\log 48}$

$\Rightarrow 1+abc=2\frac{\log 24}{\log 36}\times \frac{\log 36}{\log 48}$

$\Rightarrow 1+abc=2{\log }_{36}\left(24\right){\log }_{48}\left(36\right)$

$\Rightarrow 1+abc=2bc$