The two logs in ln(x − 2) + ln(x + 3) = 1 can be combined into one:
ln(x − 2) + ln(x + 3) = ln[ (x-2) * (x+3 ) ] = 1
Then (x-2)(x+3) = e^1 = e
Perform the indicated multiplication.
x^2 + 3x - 2x - 6 = e
1x^2 + 1x - (6+e) = 0
You can use the Quadratic Formula to find the roots here.
Note that a=1, b=1 and c = (6+e).
The discriminant is b^2 - 4ac, or 1^2 - 4(1)(6+e).
Unfortunately, the discriminant is negative, indicating that you'll have two complex roots.
