<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] $$(3+2^{1/2})^x+1=6((2^{1/2}+1)^x)$$ 
$$(2^{1/2}+1)^{2x}+1=6((2^{1/2}+1)^x)$$  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
$$(2^{1/2}+1)^x=3+2\sqrt2=(\sqrt2+1)^2\Rightarrow x_1=2$$
$$(2^{1/2}+1)^x=3-2\sqrt2=(\sqrt2+1)^{-2}\Rightarrow x_2=-2$$
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