Where am I going wrong while trying to solve this logarithmic equation? $$\log _{ 0.2 }{ x } +\log _{ \sqrt { 5 }  }{ x } =\log _{ 25 }{ x } +1$$
Steps I took:
$$\log _{ \frac { 1 }{ 5 }  }{ x } +\frac { \log _{ \frac { 1 }{ 5 }  }{ x }  }{ \log _{ \frac { 1 }{ 5 }  }{ \sqrt { 5 }  }  } =\frac { \log _{ \frac { 1 }{ 5 }  }{ x }  }{ \log _{ \frac { 1 }{ 5 }  }{ 25 }  } +1$$
$$\log _{ \frac { 1 }{ 5 }  }{ x } +-\frac { \log _{ \frac { 1 }{ 5 }  }{ x }  }{ 2 } =-\frac { \log _{ \frac { 1 }{ 5 }  }{ x }  }{ 2 } +1$$
I can keep going, but this doesn't seem to lead me to the correct answer. Where did I go wrong? 

Finding a common base might be the best approach. Re-write your equation as 
$$\frac{\log x}{\log 1/5} +\frac{\log x}{\log 5^{1/2}} = \frac{\log x}{\log 5^2} +1. $$
$$\log x \left(\frac{1}{\log 1/5} +\frac{1}{\log 5^{1/2}}-\frac{1}{\log 5^2}\right) = 1 $$
Then the solution is $x = e^{1/a}$ in which $a$ is the numerical factor on the left above. As $a$ simplifies to $a = 1/\log 25$ the answer is $x = 25.$
