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.
On
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: $$ \frac{1}{2} y^2+y^2 z-\sqrt{2} yz=1 $$
On
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 $x=-1.$
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 $y=1$.