Question 404296
{{{(1/2)logx + (1/3)logy -2logz}}}



use power rule:

{{{a log b = log a^b}}}



{{{(1/2)logx = log x^(1/2)}}}


{{{(1/3)logy = log y^(1/3)}}}

{{{-2logz = log z^(-2)}}}


The the addition rule:

{{{log x^(1/2) + log y^(1/3) + log z^(-2) = log (x^(1/2)*y^(1/3)*z^(-2))}}}

And remember for powers:
 
{{{x^(1/2)= sqrt(x)}}}

{{{y^(1/3)=root(3,y)}}}

{{{z^(-2)=1/z^2}}}


so, you will have:

{{{log x^(1/2) + log y^(1/3) + log z^(-2) =log ((sqrt(x)*root(3,y)*(1/z^2)))}}}