<s>[INST] 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. 
 [/INST] 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 $\log_2 x = a$
$$ \frac{2a}{3}= a-1 \ \implies \ a=3$$
$$\log_2 x = 3\ \implies \ x=2^3 = 8$$
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