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. 

 A: $${ x }^{ \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  }=\sqrt { 8 } \\ \log _{ 2 }{ \left( { x }^{ \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  } \right) =\log _{ 2 }{ { 2 }^{ \frac { 3 }{ 2 }  } }  } \\ \left( \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \left( \frac { 1 }{ 2 } \log _{ 2 }{ x } -1 \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \text{Let } \log _{ 2 }{ x } =t \text{ and substitute.}\\ { t }^{ 2 }-2t-3=0\\ \left( t+1 \right) \left( t-3 \right) =0\\ \log _{ 2 }{ x } =-1\Rightarrow x=0.5\\ \log _{ 2 }{ x } =3\Rightarrow x=8$$
A: $$x^{\log_2(\sqrt{x})-1} = \sqrt{8}$$
$$\log_2(x^{\log_2(\sqrt{x})-1})=\log_2(\sqrt{8})$$
$$(\log_2(\sqrt{x})-1)\log_2(x)=\frac{1}{2}\log_2(8)$$
$$\frac{1}{2}\log_2(8)\Rightarrow\frac{3}{2}$$
$$\left(\frac{1}{2}\log_2(x)-1\right)\log_2(x)=\frac{3}{2}$$
$$\log_2(x)=u$$
$$\left(\frac{1}{2}u-1\right)u*2=\frac{3}{2}*2$$
$$u^2-2u-3=0$$
$$u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=3,u=\frac{2+\sqrt{(-2)^2-4*1*(-3)}}{2}=-1$$
$$\log_2(x)=3\Rightarrow x=8,\log_2(x)=-1\Rightarrow x=\frac{1}{2}$$
$$8^{\log_2(\sqrt{8})-1} = \sqrt{8},\frac{1}{2}^{\log_2(\sqrt{\frac{1}{2}})-1} = \sqrt{8}$$
A: Fill in details and be sure you understand what properties are being applied:
$$x^{\log_2\sqrt x-1}=\sqrt8\implies\left(\frac12\log_2x-1\right)\log_2x=\frac32\implies$$
$$\left(\log_2x\right)^2-2\log_2x-3=0\implies\left(\log_2x-3\right)\left(\log_2x+1\right)=0$$
Finish now the exercise...and observe the only really new thing from what you did  is the very last step
A: $$\large\log_2\sqrt{x}=t\implies x=2^{2t}$$
$$\large{2^{2t(t-1)}}=2^{\frac{3}{2}} \implies 4t^2-4t-3=0\implies t_1=-\frac{1}{2}\,,\, t_2=\frac{3}{2}$$
then
$$\large x_1=\frac{1}{2}\,\,,\,\, x_2=8$$
