Ln x + ln (x+3)=2 could someone please answer with workings

Since ln(a) + ln(b) = ln(ab), we see that:
ln(x) + ln(x + 3) = ln[x(x + 3)].

So, we have:
ln(x) + ln(x + 3) = 2
==> ln[x(x + 3)] = 2.

Then, by the definition of logarithm, this implies that:
x(x + 3) = e^2.

Expanding the left side:
x(x + 3) = e^2
==> x^2 + 3x = e^2
==> x^2 + 3x - e^2 = 0.

Then, by the Quadratic Formula:
x = [-b ± √(b^2 - 4ac)]/(2a), with a = 1, b = 3, and c = -e^2:
= {-3 ± √[3^2 - 4(1)(-e^2)]}/[2(1)]
= [-3 ± √(4e^2 + 9)]/2.

However, since ln(x) is only defined for x > 0 and [-3 - √(4e^2 + 9)]/2 < 0, we see that the solution is:
x = [-3 + √(4e^2 + 9)]/2.

I hope this helps!

There is no need to use the quadratic formula for this problem.

Solve by using the laws of logarithms then completing the square:
lnx + ln(x + 3) = 2
ln[x(x + 3)] = 2
x(x + 3) = e²
x² + 3x = e²
4x² + 12x = 4e²
(2x + 3)² - 9 = 4e²
(2x + 3)² = 9 + 4e²
2x + 3 = ±√(9 + 4e²)
2x = -3 ± √(9 + 4e²)
x = [-3 ± √(9 + 4e²)] / 2
The negative solution does not work so the only solution is x = [-3 + √(9 + 4e²)] / 2.

ln [ x ( x + 2 ) ] = 2

x ( x + 2 ) = e^2

x ² + 2x - 7.4 = 0

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ - 2 ± √ ( 4 + 29.6 ) ] / 2

x = [ - 2 ± √ ( 33.6 ) ] / 2

x = 1.9 , x = - 3.9

ln(x) + ln(x + 3) = 2
ln(x(x + 3)) = 2
x² + 3x = e²

x = (-3 +/- √(9 - 4(-e²))/2
Select a (+) value of x.