<s>[INST] 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 :)
 [/INST] $(\sqrt2+1)^2=3+2\sqrt2$
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 $y^x=3-2\sqrt2,y^x=y^{-2}\implies x=-2$
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