Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$1+xyz$

$=1+\left({\log }_{12}6\right)\left({\log }_{18}12\right)\left({\log }_{24}18\right)$

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

$=1+\frac{\log 6}{\log 24}$

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

$={\log }_{24}(6\times 24)$

$={\log }_{24}144$

$={\log }_{24}12\times 12$

$=2{\log }_{24}12$

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

$=2yz$

Hence proved