How do you express ##log 36## in terms of ##log 2## and ##log 3##?

##log36=2*(log2+log3)##
To write this expression in terms of ##log2## and ##log3## you first have to write number ##36## as a product of powers of ##2## and ##3##.
##36=6^2=(2*3)^2##
So you can write:
##log36=log(2*3)^2##
Now you can use the properties of logarythms which say rhat:
##log a^b=b*loga##
##log(a*b)=loga+logb##
After using rule ##1## you get:
##log(2*3)^2=2*log(2*3)##
Now if you use rule ##2## you get the final result:
##2*log(2*3)=2*(log2+log3)##
Now you can write the final answer:
##log36=2*(log2+log3)##