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: 
 A: 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 $2^x = 3^y = 6^z$
A: Take $$2^x =3^y =6^{-z} =\alpha $$ $$\Rightarrow x=\log_2 \alpha =\frac {\log \alpha}{\log 2}; y =\frac {\log \alpha}{\log 3}; -z=\frac {\log \alpha}{\log 6} $$ 
Can you take it from here?
A: set $$2^x=t,3^y=t,6^{-z}=t$$ then we have $$\ln(6)=-\frac{\ln(t)}{z}$$ and $z \ne 0$ thus we get
$$\ln(2)+\ln(3)=-\frac{\ln(t)}{z}$$ now plugging the given Terms in this equation
$$\frac{\ln(t)}{x}+\frac{\ln(t)}{y}=-\frac{\ln(t)}{z}$$
can you proceed?
A: So $2^x=3^y=6^w$ (where $w=\pm z$ may be confused by a typo). We can sort this without logs etc. So this is to show a method rather than resolve typos.
Then we have $2^{x-w}=3^w$. Raise this to the power $y$ to obtain $2^{y(x-w)}=3^{wy}$. Now substitute $3^y=2^x$ so that $2^{y(x-w)}=2^{wx}$.
We can equate exponents to give $yx-yw=wx$ and divide by $wxy\neq 0$ to get $\frac 1w-\frac 1x=\frac 1y$ or$$\frac 1x+\frac 1y-\frac 1w=0$$
Note that if $wxy=0$ then the sum is undefined, but the original equation is satisfied by $w=x=y=0$, so this should have been excluded as a possibility.
