Question

Find the relation between $x,y,z$, if $x={\log }_{2k}\left(k\right),y={\log }_{3k}\left(2k\right),z={\log }_{4k}\left(3k\right)$ is

• A. $xyz+1=2yz$
• B. $xy+yz+xz=2xyz$
• C. $xy+yz+xz=xyz$
• D. $xy+yz+xz=xyz+1$

Answer 1

The correct option is A $xyz+1=2yz$

Given $x={\log }_{2k}\left(k\right),y={\log }_{3k}\left(2k\right),z={\log }_{4k}\left(3k\right)$
$\Rightarrow x=\frac{\log k}{\log 2k},y=\frac{\log 2k}{\log 3k},z=\frac{\log 3k}{\log 4k},[\because {\log }_{b}a=\frac{\log a}{\log b}]$
$\therefore xyz=\frac{\log k}{\log 4k}$
$\Rightarrow xyz+1=\frac{\log k}{\log 4k}+\frac{\log 4k}{\log 4k}$
$\Rightarrow xyz+1=\frac{\log 4k^2}{\log 4k}$
$=2\frac{\log 2k}{\log 4k},[\because \log a+\log b=\log (ab\left)\right]$
$=2yz$
$\therefore xyz+1=2yz$
Hence, option A.