Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } }  }{ \log _{ 2 }{ 8 }  } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } }  }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ 2 }{ x } =3\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\frac { 3 }{ 2 } \log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\log _{ 2 }{ (x-1)^{ \frac { 3 }{ 2 }  } } $$
This method seems to be very inefficient and I don't know how I would go from here. Can someone please point me in the right direction. Hints only please. No actual solution. 

$2\log_{8}x=\log_{2}x-1$
$2\log_{2}x=3\log_{2}x-3$ based on $\log_{8}x=\frac{\log_{2}x}{\log_{2}8}=\frac{\log_{2}x}{3}$
$3=\log_{2}x$
$8=2^{3}=x$
