Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$

Question

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.

Answer 1

Hint:

Let $\sqrt { 2^{ x-1 } } =t$ then we have $${ t }^{ 2 }-3t+2=0$$

Answer 2

$$2^{x-1}-3 \sqrt{2^{x-1}}+2=0$$ $$2^{x-1}+2=3 \sqrt{2^{x-1}}$$ $$9\cdot 2^{x-1}=2^{(x-1)*2}+2^{x+1}+4$$ $$9\cdot2^x\cdot2^{-1}=2^{2x-2}+2^x\cdot2+4$$ $$9\cdot2^x\cdot\frac{1}{2}=2^{2x}\cdot2^{-2}+2^x\cdot2+4$$ $$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{2^2}+2^x\cdot2+4$$ $$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{4}+2^x\cdot2+4$$ 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 $$x=1$$ $$x=3$$