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

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:

$\ln \left(a\right) + \ln \left(b\right) = \ln \left(a \cdot b\right)$

Substituting $x$ for $a$ and $x + 3$ for $b$ we get

$\ln \left(x \left(x + 3\right)\right) = 1$

Now we take the $e$-xponential of both sides, which gives us

$x \left(x + 3\right) = e$

Multiplying out the left side, we get

${x}^{2} + 3 x = e$

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:

${\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$

We now have

x^2+3x + ? = e + ?

${x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} = e + {\left(\frac{3}{2}\right)}^{2}$

Rewriting the left-hand side using the factor $\frac{3}{2}$ we now have

${\left(x + \frac{3}{2}\right)}^{2} = e + {\left(\frac{3}{2}\right)}^{2}$

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 $\frac{3}{2}$ 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