Question

If $x={\log }_{12}6,y={\log }_{18}12$ and $z={\log }_{24}18$ prove that $1+xyz=2yz$

Answer 1

$=\frac{\log 6}{\log 12};y={\log }_{8}12=\frac{\log 12}{\log 18}$

$z={\log }_{24}18=\frac{\log 18}{\log 24}$

$\Rightarrow xyz=\frac{\log 6}{\log 12}\times \frac{\log 12}{\log 18}\times \frac{\log 18}{\log 24}=\log 6\times \frac{1}{\log 18}\times \frac{\log 18}{\log 24}$

$xyz=\frac{\log 6}{\log 24}$

Now

$1+xy=\frac{\log 24+\log 6}{\log 24}=\frac{\log (6\times 4)+\log 6}{\log 24}$

$=\frac{\log 6+2\log 2+\log 6}{\log 24}=\frac{2[\log 2+\log 6]}{\log 24}$

$=2\frac{\log 122}{\log 24}=2\frac{\log 12}{\log 24}\times \frac{\log 18}{\log 24}=2yz$