Question 993985
if
{{{log(3)=a}}} 
{{{log(4)=b}}}=>{{{log(2^2)=b}}}=>{{{2log(2)=b}}}=>{{{log(2)=b/2}}}

then 
{{{log((sqrt(18)))}}}

={{{log((sqrt(2*9)))}}}

={{{log((3sqrt(2)))}}}

={{{log(3)+log((2^(1/2)))}}}

={{{log(3)+(1/2)log(2)}}}...since {{{log(3)=a}}} and {{{log(2)=b/2}}}, we have


={{{a+(1/2)(b/2)}}}

={{{a+b/4}}}