equation with exponential functions 3 
Solve the following equation over the real numbers(without utilising calculus)
  $$ (\frac{1}{2})^{1+x}+(\frac{1}{6})^{x}-\sqrt{2}\cdot(\frac {\sqrt{2}}{6})^x=1 $$

I simply can't really find a solid starting point; I thinks it might be a notation tham I'm oblivious to; I do know the root of the equation $ x=-1 $ thanks to Wolfram, but am interested more in the steps to determine it.
 A: To get rid of all the fractions, set $y=-x$ to get
$$2^{y-1}+6^y-\sqrt{2}(3\sqrt{2})^y=1,$$
from which it is quickly clear that $y=1$ is a solution. Moreover, rearranging shows that 
$$1-2^{y-1}=6^y-\sqrt{2}(3\sqrt{2})^y=3^y(2^y-\sqrt{2}^{y+1})\tag{1},$$
and if $y\neq1$ then $2^y-\sqrt{2}^{y+1}\neq0$ and so we find that
$$3^y=\frac{1-2^{y-1}}{2^y-\sqrt{2}^{y+1}}=-\frac{1}{2}-\sqrt{2}^{-y-1},$$
where the left hand side is positive and the right hand side is negative, a contradiction. Hence there is no solution with $y\neq1$, i.e. the unique solution is $y=1$.
A: Hint:
First rewrite as:
$$
\frac{1}{2}\left(\frac{1}{2}\right)^{x}+\left(\frac{1}{6}\right)^{x}-\sqrt{2}\cdot\left(\frac {\sqrt{2}}{6}\right)^x=1 
$$
Now the idea is to replace the exponentials with other variables:
$$
y=\left(\frac {1}{\sqrt{2}}\right)^x
$$
$$
z=\left(\frac {1}{3}\right)^x
$$
Then:
$$
\frac{1}{2} y^2+y^2 z-\sqrt{2} yz=1
$$
A: We need to solve that
$$\frac{3^{x}}{2}+1-(\sqrt2)^{x+1}=6^x$$ or
$$3^x\left(2^x-\frac{1}{2}\right)+(\sqrt2)^{x+1}-1=0$$ or
$$3^x\left((\sqrt2)^{x+1}-1\right)\left((\sqrt2)^{x+1}+1\right)+2\left((\sqrt2)^{x+1}-1\right)=0$$ or
$$\left((\sqrt2)^{x+1}-1\right)\left(3^x\left((\sqrt2)^{x+1}+1\right)+2\right)=0$$ or
$$(\sqrt2)^{x+1}=1,$$ which gives $x=-1.$
