Question

If $x={\log }_{24}12,$ $y={\log }_{36}24$ and $z={\log }_{48}36,$ then $(1+xyz)$ equals $k$ times $yz.$ The value of $k$ is

Answer 1

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