If $2^x=3^y=6^{-z}$ then find:$ \frac{1}{x}+\frac{1}{y}-\frac{1}{z}$ I am a grade 8 student and i am aware that $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ is proved by setting all values $= k$.
See the question number 24 in my assignment.    
1: 

Actually you have because of misprint mistake in your question.
$2^x = 3^y = 6^{-z} = k $
$ 2 = k^{\frac{1}{x}}$
$ 3 = k^{\frac{1}{y}}$
$ 6 = k^{\frac{1}{-z}}$
As we know,
2.3 = 6 
$k^{\frac{1}{x}} \cdot k^{\frac{1}{y}} = k^{\frac{1}{-z}}$
$k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{-z}}$
On comparing powers we have,
$\frac{1}{x} + \frac{1}{y} = \frac{1}{-z}$
$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

To prove $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 0$

We should have expression $2^x = 3^y = 6^z$
