How do you solve #Log(x+2)+log(x-1)=1#?
Using the addition property of logs you know that
this equals log(x+2)(x-1)=1
and using the common log you know that (x+2)(x-1) needs to equal 10 for the equation to be true, (x+2)(x-1)=10 and
this evaluates to (x+4)(x-3)=0, but -4 can't be a solution because that would make one of the original logs undefined.