Question

If $2^x=3^y=6^{-z}$, find the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer 1

Find the value of the required expression.

Let, $2^x=3^y=6^{-z}=k\left(\text{say}\right)$

$\Rightarrow 2=k^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.},3=k^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.},6=k^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$z$}\right.}$

$\begin{array}{l}\because 6=2\times 3=k^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\times k^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}=k^{\frac{1}{x}+\frac{1}{y}\left[\text{Using above results}\right]}\\ \Rightarrow k^{-\frac{1}{z}}=k^{\frac{1}{x}+\frac{1}{y}}\\ \Rightarrow -\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\end{array}$

Hence, the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is $0$.