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:

1. ##log a^b=b*loga##
2. ##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)##