Question

If $x,y,z$ are positive real numbers such that ${\log }_{2x}z=3,{\log }_{5y}z=6$ and ${\log }_{xy}z=\frac{2}{3}$, then the value of $\frac{1}{2z}$ is

Answer 1

$\frac{lnz}{ln\left(2x\right)}=3$ and $\frac{lnz}{ln5y}=6$
Or
$\frac{ln2x}{lnz}+\frac{ln5y}{lnz}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$.
Or
$\frac{ln\left(2x\right)+ln\left(5y\right)}{ln\left(z\right)}=\frac{1}{2}$
Or
$\frac{ln\left(10xy\right)}{lnz}=\frac{1}{2}$
Or
$\frac{ln10}{lnz}+\frac{ln\left(xy\right)}{lnz}=\frac{1}{2}$
Or
$\frac{ln10}{lnz}+\frac{3}{2}=\frac{1}{2}$
Or
$\frac{ln10}{lnz}=-1$
$ln\left(z\right)=\frac{1}{10}$
$z=\frac{1}{10}$
Hence
$\frac{1}{2z}=\frac{10}{2}=5$