Question

If $a^x=b^y=c^z$ and $b^2=ac$, find $\frac{1}{x}+\frac{1}{z}-\frac{2}{y}$

Answer 1

Taking log we get

$x\log a=y\log b=z\log c=k\left(Let\right)$

and $b^2=ac\Rightarrow \frac{ac}{b^2}=1$

Now, $\frac{1}{x}+\frac{1}{z}-\frac{2}{y}=\frac{\log a}{k}+\frac{\log c}{k}-\frac{2\log b}{k}=\frac{1}{k}.(\log a+\log c-\log b^2)=\frac{1}{k}.\left(\log \right(\frac{ac}{b^2}\left)\right)=\frac{1}{k}.\log 1=0$