Question 107094
Let's look at what we know about logarithms:

{{{loga(x^n)=n*loga(x)}}},


{{{loga(x) + loga(y) = loga(x*y)}}}


{{{loga(x) - loga(y) = loga(x/y)}}}

Using these facts, let's operate on your orignial expression:

{{{(1/2)loga(x) + 4*loga(y) -(1/3)loga(z)}}}


Beginning with the first term, remember that {{{x^(1/2)=sqrt(x)}}}, so


{{{(1/2)loga(x)=loga(sqrt(x))}}}


Similarly, the other two terms become:


{{{4*loga(y)=loga(y^4)}}}, and 
{{{(1/3)loga(z) = loga(z^(1/3))}}}

({{{z^(1/3)}}} is the cube root of z, but I can't render that on this system)


Putting the terms together, using the second rule from above:


{{{loga((y^4)*sqrt(x)/z^(1/3))}}}


The answer would be much neater if I could replace the fractional power on z with a cube root radical, but it is still correct in this form.