Exponential equation, $(3+2\sqrt2)^x+1=6(\sqrt2+1)^x$ http://i.stack.imgur.com/YP2Ha.png
$$(3+2\sqrt2)^x+1=6(\sqrt2+1)^x \qquad\qquad x\in\mathbb{R}$$ 
I managed to find one of the solutions (x=2), but I got stuck. I would really appreciate a step by step solution. Thanks in advance :)
 A: $(\sqrt2+1)^2=3+2\sqrt2$
If $\sqrt2+1=y, y^{2x}+1=6{y^x}\implies y^{2x}-6y^x+1=0, y^x=\frac{6\pm\sqrt{6^2-4}}2=3\pm2\sqrt2$
Now, $3+2\sqrt2=y^2\implies 3-2\sqrt2=\frac1{3+2\sqrt2}=y^{-2}$
If $y^x=3+2\sqrt2,y^x=y^2\implies x=2$ as $y=\sqrt2+1\ne0,\pm1$
Similarly for $y^x=3-2\sqrt2,y^x=y^{-2}\implies x=-2$
A: $$(3+2^{1/2})^x+1=6((2^{1/2}+1)^x)$$ 
$$(2^{1/2}+1)^{2x}+1=6((2^{1/2}+1)^x)$$  taking$(2^{1/2}+1)^x=y$ we get
$$y^2-6y+1=0$$ from there
$$y_{1,2}=\frac{6\pm\sqrt{32}}{2}=\frac{6\pm4\sqrt{2}}{2}=3\pm 2\sqrt2$$ or
$$(2^{1/2}+1)^x=3+2\sqrt2=(\sqrt2+1)^2\Rightarrow x_1=2$$
$$(2^{1/2}+1)^x=3-2\sqrt2=(\sqrt2+1)^{-2}\Rightarrow x_2=-2$$
A: $\begin{eqnarray} 
{\bf Hint}\ &&\rm\ \  f(x) &=\,&\rm\ (x-\alpha^2)\ (x-\alpha^{-2}) \\
&&  &=\,&\rm \ \ x^2- (b^2\!-2)\,x\, +\, 1,\ \ b\, =\, \alpha + \alpha^{-1} \\ 
\rm Thus\ \ &&\rm f(\alpha^n) &=&\rm \alpha^{2n}\!- (b^2\!-2)\,\alpha^n\! + 1 \\
 &&\rm  &=&\rm (\alpha^n-\alpha^2\,)\ (\,\alpha^n-\alpha^{-2})\, = \,0 \\
 &&\iff &&\rm\ \alpha^n\! =\rm \alpha^2\,\  or  \,\ \alpha^n\! = \alpha^{-2}
\end{eqnarray}$
Your is $\rm\,\ \alpha=\sqrt{2}+\!1,\ \alpha^{-1}\!=\sqrt{2}-\!1,\ b = \alpha\!+\!\alpha^{-1}\! = 2\sqrt{2},\,\ f(x) = x^2\!-6x+1$ 
$\rm\quad so\quad \alpha^{2n}\!+\!1\, =\, (3\!+\!2\sqrt{2})^n\! + 1\, =\, 6\,(\sqrt{2}\!+\!1)^n =\, (b^2\!-\!2)\, \alpha^n\!\iff f(\alpha^n) = 0$
