2021/093) Given $2^x = 3^y = 6^{-z}$ evaluate $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}$

we are given $2^x = 3^y = 6^{-z}$

so $2 = 6^{\frac{-z}{x}}$

and $3 = 6^\frac{-z}{y}$

so $2 * 3 = 6^{(-\frac{z}{x} - \frac{z}{y})}$

or $6^1 = 6^{(-\frac{-z}{x} - \frac{z}{y})}$

or $1 = -\frac{z}{x} - \frac{z}{y}$

or  $\frac{z}{x} +\frac{z}{y} + 1 = 0 $

or $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}= 0$