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. 

$$
\log_{2} 2x = \log_{4} 4x^{6} - \log_{2}2x \quad \text{iff} \quad \frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2};\\
\frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2} \quad \text{iff} \quad 2\log 2x = \log 4x^{6} - 2\log 2x;\\
2\log 2x = \log 4x^{6} - 2\log 2x \quad \text{iff} \quad
\log (2x)^{4} = \log 4x^{6};\\
\log (2x)^{4} = \log 4x^{6} \quad \text{iff} \quad x = 2.
$$
