Finding a solution to this logarithm equation. How is: 
$ \log_{(xyz)^{1/3}} \left(\frac{yz}{x^{3}}\right)^{2}$
expressed in terms of a and b when:
$ a = \log_{y} x $
$ b = \log_{z} x $
EDIT: It is z in the numerator, apologies i posted by memory. Also reframed the question.
 A: $$\log _{ (xyz)^{ \frac { 1 }{ 3 }  } } \left( \frac { yz }{ x^{ 3 } }  \right) ^{ 2 }=\frac { 2 }{ \frac { 1 }{ 3 }  } \log _{ (xyz) } \left( \frac { yz }{ x^{ 3 } }  \right) =6\frac { \log _{ x }{ \left( \frac { yz }{ x^{ 3 } }  \right)  }  }{ \log _{ x }{ \left( xyz \right)  }  } =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right)  } -3\log _{ x }{ x }  }  }{ \log _{ x }{ \left( x \right) +\log _{ x }{ \left( y \right)  } +\log _{ x }{ \left( z \right)  }  }  } =\\ =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right)  } -3 }  }{ 1+\log _{ x }{ \left( y \right)  } +\log _{ x }{ \left( z \right)  }  } =6\frac { \frac { 1 }{ \log _{ y }{ \left( x \right)  }  } +\frac { 1 }{ \log _{ z }{ \left( x \right)  }  } -3 }{ 1+\frac { 1 }{ \log _{ y }{ \left( x \right)  }  } +\frac { 1 }{ \log _{ z }{ \left( x \right)  }  }  } =6\frac { \frac { 1 }{ a } +\frac { 1 }{ b } -3 }{ 1+\frac { 1 }{ a } +\frac { 1 }{ b }  } =6\frac { \frac { a+b-3ba }{ ba }  }{ \frac { ab+a+b }{ ab }  } =\\ =\frac { 6a+6b-18ba }{ ab+a+b } $$
A: This can be written as, 
$$\frac{\log (\frac{yz}{x^3})^2}{\log (xyz)^{1/3}}$$
=$$\frac{6\log y+6\log z-18\log x}{\log x+ \log y + \log z}$$
Now given that $$a=\frac {\log x}{\log y},b=\frac{\log x}{log z}$$
we get $\log x=a \log y $ and $\log z= \frac{a}{b}\log y$
Now simply substitute the values of $\log x$ and $\log z$ in the expression to get
$$\frac{6\log y+6\frac{a}{b}\log y- 18a \log y}{a\log y+\log y+\frac{a}{b}\log y}$$
=$$\frac{6b+6a-18ab}{ab+b+a}$$
