Discussion Forum

If $2^{x}=3^{y}=6^{-z}$, then $\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is equal to?

Answer:

Explanation:

Let $2^{x}=3^{y}=6^{-z}=k$ $\Leftrightarrow 2=k^{\frac{1}{x}},$ $\Leftrightarrow 3=k^{\frac{1}{y}}$ and $\Leftrightarrow 6=k^{\frac{-1}{z}}.$ Now $2\times 3=6$ $\Leftrightarrow k^{\frac{1}{x}} \times k^{\frac{1}{y}}=k^{\frac{- 1}{z}}$ $\Leftrightarrow k^{\left(\frac{1}{x} +\frac{1}{y} \right)}=k^{\frac{- 1}{z}}$ $\therefore \frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$ $\Leftrightarrow \frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$