Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$ Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$
It seemed easy for me, but I couldn't do it. By Wolfram Alpha, I got that $x=1,3$, but I don't know how to get there.
Any hints?
Thanks.

$$2^{x-1}-3 \sqrt{2^{x-1}}+2=0$$
$$2^{x-1}+2=3 \sqrt{2^{x-1}}$$
$$9\cdot 2^{x-1}=2^{(x-1)*2}+2^{x+1}+4$$
$$9\cdot2^x\cdot2^{-1}=2^{2x-2}+2^x\cdot2+4$$
$$9\cdot2^x\cdot\frac{1}{2}=2^{2x}\cdot2^{-2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{2^2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{4}+2^x\cdot2+4$$
Set $ n=2^x$
$$\frac{9}{2}n=n^2\cdot\frac{1}{4}+n\cdot2+4$$
$$\frac{9}{2}n=\frac{n^2}{4}+2n+4$$
$$0=\frac{n^2}{4}-\frac{5}{2}n+4$$
$$0=n^2-10n+16$$
Factor:$$(n-8)(n-2)=0$$
$$n=2, 8$$
Since $n=2^x$, we get
$$2^x=2$$
$$2^x=8$$
Solving these, we get 
$$x=1$$
$$x=3$$
