Question 131513
ln(x+2)+ln(x-2)=0 
ln[(x+2)(x-2)] = 0
ln[x^2-4] = 0
x^2-4 = e^0
x^2-4 = 1
x^2 = 5
x = sqrt5 or x = -sqrt(5)
Only x = sqrt5 is an acceptable answer
because -sqrt(5)-2 is negative and ln(negative) does not exist.
======================
Cheers,
Stan H.