Question 200475
Apply log rules:
log (x) + log ( x + 3 ) = 1
log [(x)(x+3)] = 1
(x)(x+3) = 10^1
x^2 + 3x = 10
x^2 + 3x - 10 = 0
(x+5)(x-2) = 0
x = {-5, 2}
Throwing out the negative solution we are left with:
x = 2