Question

If $a=lo{g}_{24}12,b=lo{g}_{36}24,c=lo{g}_{48}36$. Prove that $1+abc=2bc$.

Answer 1

$a={\log }_{24}^{12}$ $b={\log }_{36}^{24}$ $c={\log }_{48}^{36}$

$abc={\log }_{24}^{12}{\log }_{36}^{24}{\log }_{48}^{36}$

$=\frac{\log 12}{\log 24}\frac{\log 24}{\log 36}\frac{\log 36}{\log 48}$

$ab={\log }_{48}^{12}$

$1+abc={\log }_{48}^{48}+{\log }_{48}^{12}$ $\log ab=\log a+\log b$

$={\log }_{48}^{48\times 12}={\log }_{48}^{12\times 12\times 4}={\log }_{48}^{{24}^2}=2{\log }_{48}^{24}$

$2bc=2\times {\log }_{36}^{24}\times {\log }_{48}^{36}$

$=2{\log }_{48}^{24}$

$\therefore 1+abc=2bc$ .