Question 1020929
{{{log(3,a) = b}}} ==> {{{log(a,3) = 1/b}}}

{{{log(a, 2) = c}}} ==> {{{log(a,16) = 4log(a,2) = 4c}}}

==> {{{log(a,48) = log(a,3*16) = log(a,3) + log(a,16) = 1/b+4c}}}

Problem solved.