<s>[INST] Solving $(\sqrt{2})^x+(\sqrt{2})^{x-1}=2(2\sqrt{2}+1)$ I'm in stuck with this simple equation.
$$(\sqrt{2})^x+(\sqrt{2})^{x-1}=2(2\sqrt{2}+1)$$
This is my solution:
$$\begin{align}(\sqrt{2})^x+(\sqrt{2})^x(\sqrt{2})^{-1} &=4\sqrt{2}+2 \tag{1}\\[4pt]
2^{x/2}+2^{(x-1)/2}&=2^2\cdot 2^{1/2}+2^1 \tag{2}\\[4pt]
{\frac x2}+{\frac {x-1} 2}&=2^{2+\frac 1 2}+2^1 \tag{3}\\[4pt]
{\frac {2x-1} 2}&={\frac 5 2}+1  \tag{4}\\[4pt]
{\frac {2x-1} 2}&={\frac 7 2} \tag{5}\\[4pt]
{\frac {2x-1} 2}-{\frac 7 2}&=0 \tag{6}\\[4pt]
{\frac {2x-8} 2}&=0 \tag{7}\\[4pt]
2x&=8 \tag{8}\\[4pt]
x&=4 \tag{9}
\end{align}$$
But the solution must be $x=4.33$. I can't find the error.
Is here someone who can help me? Thanks
 [/INST] Your incorrect step starts from here.
$$2^{\frac x2}+2^{\frac{x-1}{2}}≠\frac x2+\frac{x-1}{2}$$
Because, this is not a equality:
$$2^a+2^b ≠ a+b$$

You can do one of the correct solutions as follows:
$$(\sqrt 2)^{x-1}(\sqrt 2-1)=4\sqrt 2+2$$
$$(\sqrt 2)^{x-1}=\frac{4\sqrt 2+2}{\sqrt 2-1}$$
$$(\sqrt 2)^{x-1}=\frac{(4\sqrt 2+2)(\sqrt 2+1)}{(\sqrt 2-1)(\sqrt 2+1)}$$
$$(\sqrt 2)^{x-1}=(4\sqrt 2+2)(\sqrt 2+1)$$
$$(\sqrt 2)^{x-1}=6\sqrt 2+10$$
$$\frac {x-1}{2}=\log_2(6\sqrt 2+10)$$
$$\cdots \cdots \cdots$$
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