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

1 Answer
Jul 14, 2016

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:

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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