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.
Using a change of base: $$\log_b a = \frac{\log_n a}{\log_n b}$$

Change $\log_8 x^2$ base to 2: $$\log_8 x^2 = \frac{\log_2 x^2}{\log_2 8} = \frac{2\log_2 x}{3} $$

Let $\log_2 x = a$

$$ \frac{2a}{3}= a-1 \ \implies \ a=3$$

$$\log_2 x = 3\ \implies \ x=2^3 = 8$$
$\log_a(x)=\frac{\ln(x)}{\ln(a)}$

$2\log_8(x)=\log_2(x)-1$

$2\frac{\ln(x)}{3\ln(2)}=\frac{\ln(x)}{\ln(2)}-1$

$-\frac{\ln(x)}{3\ln(2)}=-1$

$\frac{1}{3}\log_2(x)=1$

$\log_2(x)=3$

$x=8$