Question

If $2^x=3^y=6^z$, Prove that $(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0)$

Answer 1

$2^x=3^y=6^z=k$

$x={\log }_{2}ky={\log }_{3}kz={\log }_{6}k$

$\frac{1}{x}={\log }_{k}2\frac{1}{y}={\log }_{k}3\frac{1}{z}={\log }_{k}6$

$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}={\log }_{k}2+{\log }_{k}3-{\log }_{k}6$

$={\log }_{k}6-{\log }_{k}6$

$=0$