Question

If $x={\log }_{24}12,$ $y={\log }_{36}24$ and $z={\log }_{48}36,$ then $(1+xyz)$ equals

• A. $2$
• B. $2yz$
• C. $2xz$
• D. $2xy$

Answer 1

The correct option is B $2yz$

$1+xyz=1+\frac{\log 12}{\log 24}\cdot \frac{\log 24}{\log 36}\cdot \frac{\log 36}{\log 48}$
$=1+\frac{\log 12}{\log 48}$
$=\frac{\log (48\times 12)}{\log 48}=\frac{\log (24{)}^2}{\log 48}=\frac{2\log 24}{\log 48}$
$=2\cdot \frac{\log 24}{\log 36}\cdot \frac{\log 36}{\log 48}$
$=2yz$