How do you solve #lnx+ln(x+3)=1#?

1 Answer
Jul 14, 2016

Answer:

#x = sqrt(e + 9/4) - 3/2 ≈ 0.72896#

Explanation:

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:

www.wolframalpha.com

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#