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. 

$\log_a(x)=\frac{\ln(x)}{\ln(a)}$
$2\log_8(x)=\log_2(x)-1$
$2\frac{\ln(x)}{3\ln(2)}=\frac{\ln(x)}{\ln(2)}-1$
$-\frac{\ln(x)}{3\ln(2)}=-1$
$\frac{1}{3}\log_2(x)=1$
$\log_2(x)=3$
$x=8$
