Stuck solving a logarithmic equation $$\log _{ 2 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } -\log _{ 2 }{ 2x } $$
Steps I took:
$$\frac { \log _{ 4 }{ 2x }  }{ \log _{ 4 }{ 2 }  } =\log _{ 4 }{ 4x^{ 6 } } -\frac { \log _{ 4 }{ 2x }  }{ \log _{ 4 }{ 2 }  } $$
$$2\log _{ 4 }{ 2x } +2\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
$$4\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
At this point I get stuck I don't think turning this into $\log _{ 4 }{ (2x)^{ 4 } } =\log _{ 4 }{ 4x^{ 6 } } $ is the right answer. It leads to: $16x^{ 4 }=4x^{ 6 }$ and this has what seem to be extraneous solutions. 

Using: $\log_a b=\frac{\log a}{\log b}, \log ab=\log a + \log b, \log_a a=1$
$$\implies\log _{ 2 }2+\log _{ 2 }{ x } =\log _{ 4 }{ 4}+\log _{ 4 }{ x^{ 6 } } -\log _{ 2 }{ 2}-\log _{ 2 }{ x } $$
$$\implies1+\log _{ 2 }{ x } =6\frac{\log _{ 2 }{ x }}{\log_2 4} -\log _{ 2 }{ x } +1-1$$
$$\implies1+\log _{ 2 }{ x } =6\frac{\log _{ 2 }{ x }}{2}-\log _{ 2 }{ x } $$
$$\implies1+\log _{ 2 }{ x } =3{\log _{ 2 }{ x }}-\log _{ 2 }{ x } $$
$$\implies{\log _{ 2 }{ x }}=1 $$
$$\implies x=2^1$$
