<s>[INST] Solving an equation with logarithms: $x^{\log_2(\sqrt{x})-1} = \sqrt{8}$ The equation I'm given is
$$\large x^{\log_2(\sqrt{x})-1} = \sqrt{8}$$
I've tried on solving it and my best try is on the photo. Got stuck there and not sure how to proceed any further. 

 [/INST] $$x^{\log_2(\sqrt{x})-1} = \sqrt{8}$$
$$\log_2(x^{\log_2(\sqrt{x})-1})=\log_2(\sqrt{8})$$
$$(\log_2(\sqrt{x})-1)\log_2(x)=\frac{1}{2}\log_2(8)$$
$$\frac{1}{2}\log_2(8)\Rightarrow\frac{3}{2}$$
$$\left(\frac{1}{2}\log_2(x)-1\right)\log_2(x)=\frac{3}{2}$$
$$\log_2(x)=u$$
$$\left(\frac{1}{2}u-1\right)u*2=\frac{3}{2}*2$$
$$u^2-2u-3=0$$
$$u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=3,u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=-1$$
$$\log_2(x)=3\Rightarrow x=8,\log_2(x)=-1\Rightarrow x=\frac{1}{2}$$
$$8^{\log_2(\sqrt{8})-1} = \sqrt{8},\frac{1}{2}^{\log_2(\sqrt{\frac{1}{2}})-1} = \sqrt{8}$$
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