Question 143557
{{{log(b,x)=y}}} means {{{b^y=x}}}  and {{{log(b,xy)=log(b,x)+log(b,y)}}}, so


{{{log(sqrt(2),2sqrt(2))=log(sqrt(2),2)+log(sqrt(2),sqrt(2))}}}


if {{{y=log(sqrt(2),2)}}}, then {{{(sqrt(2))^y=2}}}.  If {{{(sqrt(2))^y=2}}} then {{{y=2}}}


if {{{z=log(sqrt(2),sqrt(2))}}}, then {{{(sqrt(2))^z=sqrt(2)}}}.  If {{{(sqrt(2))^z=sqrt(2)}}}, then {{{z=1}}}  (In fact, {{{log(b,b) = 1}}}, always)


Therefore:   {{{log(sqrt(2),2sqrt(2))=log(sqrt(2),2)+log(sqrt(2),sqrt(2))=2 + 1 = 3}}}