# How do you solve ln(x - 3) + ln(x + 4) = 1?

Jul 14, 2015

$x \cong 3.37$

#### Explanation:

First thing to note is that a $\ln \left(x - 3\right)$ is defined only when $x - 3 > 0 \implies x > 3$

And also $\ln \left(x + 4\right)$ is defined when $x + 4 > 0 \implies x > - 4$

And so for both functions to be defined we need to intersect their domains, leading to : $x > 3$

Now, that we have the domain we can solve the equation in this domain, thus:

$\ln \left(x - 3\right) + \ln \left(x + 4\right) = 1$

We apply the logarithm laws $\implies \ln \left[\left(x - 3\right) \left(x + 4\right)\right] = 1$

$\implies \left(x - 3\right) \left(x + 4\right) = e$

$e$ is Euler's constant

$\implies {x}^{2} + x - 12 = e$

$\implies {x}^{2} + x - 12 - e = 0$

$\implies x = \frac{- 1 \pm \sqrt{1 - 4 \left(- 12 - e\right)}}{2} = \frac{- 1 \pm \sqrt{49 + 4 e}}{2}$
The solution $\frac{- 1 - \sqrt{49 + 4 e}}{2}$ has to be rejected because it is negative and so falls out our domain $x > 3$
The only solution is $x = \frac{- 1 + \sqrt{49 + 4 e}}{2} \cong \textcolor{b l u e}{3.37}$