For this problem, we can use a property of logarithms, one of which is the following:
#ln(a)+ln(b) = ln(a*b)#
Substituting #x# for #a# and #x+3# for #b# we get
#ln( x(x+3)) = 1#
Now we take the #e#-xponential of both sides, which gives us
#x(x+3) = e#
Multiplying out the left side, we get
#x^2+3x = e#
The following step requires us to complete the square. In order to do so, we take the coefficient on the #x#-term, namely #3#, divide it by #2#, and square the result. This new number should then be added to both sides of the equation in the following way:
#(3/2)^(2) = 9/4#
We now have
#x^2+3x + ? = e + ?#
#x^2 + 3x + (3/2)^(2) = e + (3/2)^(2)#
Rewriting the left-hand side using the factor #3/2# we now have
#(x+3/2)^2 = e + (3/2)^(2)#
Taking the square root of both sides gives us
#(x+3/2) = ± sqrt(e + (3/2)^(2))#
Isolating the #x# on the left by subtracting #3/2# from both sides then yields
#x = ± sqrt(e + (3/2)^(2)) - 3/2#
Simplifying even further gives us
#x = ± sqrt(e + 9/4) - 3/2#
So the solutions are
#x = sqrt(e + 9/4) - 3/2 ≈ 0.72896#
#x = - sqrt(e + 9/4) - 3/2 ≈ -0.72896#
Which one do we keep?
Let's graph both equations:
As we can see, #x ≈ -0.72896# does not even touch the original curve, and so we only keep the positive solution, namely
#x = sqrt(e + 9/4) - 3/2 ≈ 0.72896#
