Rewrite the expression
ln(a+b)+3ln(ab)4lnc\displaystyle {\ln{{\left({a}+{b}\right)}}}+{3}{\ln{{\left({a}-{b}\right)}}}-{4}{\ln{{c}}}
as a single logarithm lnA\displaystyle {\ln{{A}}}. Then the function
A=\displaystyle {A}=