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. 


$${ x }^{ \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  }=\sqrt { 8 } \\ \log _{ 2 }{ \left( { x }^{ \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  } \right) =\log _{ 2 }{ { 2 }^{ \frac { 3 }{ 2 }  } }  } \\ \left( \log _{ 2 }{ \left( \sqrt { x }  \right) -1 }  \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \left( \frac { 1 }{ 2 } \log _{ 2 }{ x } -1 \right) \log _{ 2 }{ x } =\frac { 3 }{ 2 } \\ \text{Let } \log _{ 2 }{ x } =t \text{ and substitute.}\\ { t }^{ 2 }-2t-3=0\\ \left( t+1 \right) \left( t-3 \right) =0\\ \log _{ 2 }{ x } =-1\Rightarrow x=0.5\\ \log _{ 2 }{ x } =3\Rightarrow x=8$$
