Question 891923
Answer is {{{2a-3}}}.
Tricky, but here is the trick.......


{{{log(2,12)=a}}}
{{{a=log(2,2*2*3)}}}
{{{log(2,((2^2)*3))}}}
{{{log(2,2^2)+log(2,3)}}}
{{{2*log(2,2)+log(2,3)}}}
{{{2+log(2,3)=a}}}
{{{highlight_green(log(2,3)=a-2)}}}---This is what you will use.


Now Starting at {{{log(2,18)}}},
{{{log(2,(2*3^2))}}}
{{{log(2,2)+log(2,(3^2))}}}
{{{1+2*log(2,3)}}}-----You simply substitute what you just found for {{{log(2,3)}}} !
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