If $2^x=3^y=6^{-z}$ then prove that:$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$
I have tried starting with taking logartithms, but that gives just some more equations.
Any specific way to solve these type of problems?
Any help will be appreciated.
 A: $$2^x = 3^y = 6^{-z} = k $$
so$$x = \log_2k$$
$$ y = \log_3k$$
$$z= -\log_6k$$
so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k2 + log_k3 -\log_k6$$
 $$=\log_k{\frac{2\times3}{6}}$$
     $$=0$$
A: Take logarithms (any base) to obtain $$x\log 2= y\log 3=-z\log 6=-z\log 3-z\log 2$$
Then note that $(y+z)\log 3=-z\log 2$ and $x\log 2 = y\log 3$
Multiply left- and right- hand sides to obtain $$(xy+yz)\log 3 \log 2=-yz\log 2\log 3$$
Whence $xy+yz+zx=0$
Note that we have done no division so far, except by the non-zero term $\log 2\log 3$, and also that $x=y=z=0$ is a solution. If $xyz\neq 0$ we can divide by $xyz$ to obtain the required equation.
A: As $\log_aa=1$ and $\log_a(bc)=\log_ab+\log_ac,$
applying logarithm wrt $2,$ 
$x=y\log_23=-z\log_26=-z(1+\log_23)$
$x=y\log_23\implies \log_23=\frac xy$
and put this value of  $\log_23$   in  $\displaystyle x=-z(1+\log_23)$ 
to eliminate $\log_23$ and simplify.
A: $2^x=3^y=6^{-z}=k $ say, then $2= k^{1\over x},3=k^{1\over y},6=k^{-1\over z}$ now can you go on?
then $k^{-1\over z}=6=2\times 3 = k^{1\over x}\times k^{1\over y}=k^{{1\over x}+{1\over y}}$
A: My answer does not solve the problem at hand (because the existing solutions work fine) but it addresses an issue related to the condition $x,y,z \neq 0$.
It may seem that $x = y = z = 0$ is the only solution to the above system of equations and by imposing the condition $x,y,z \neq 0$, we are driving even that out thereby leaving no real solution to it. But it is not true.

If $x,y,z \neq 0$, then too, this system of equations has infinitely many solutions. 
In the above figure, three distinct values of $x, y, z$ are marked as a solution where $2^x = 3^y = 6^{-z} =v$. We can slide the line of $v$ up or down thereby getting infinite solutions.   
