# How do you solve  lnx+ln(x-2)=3?

We use the fact that $\ln a + \ln b = \ln a b$

#### Explanation:

Hence we have that

lnx+ln(x-2)=3=>ln(x(x-2))=3=> x(x-2)=e^3=> x^2-2x-e^3=0

The last is a trinomial with respect to x that has the following solutions

${x}_{1} = 1 - \sqrt{1 + {e}^{3}} , {x}_{2} = 1 + \sqrt{1 + {e}^{3}}$

Now we must not forget that $x > 2$ so the only acceptable solution here is $1 + \sqrt{1 + {e}^{3}}$

Sep 10, 2015

Try this:

#### Explanation:

We can use the rule of the logs that says:
$\log x + \log y = \log \left(x y\right)$ to get:

$\ln \left[x \left(x - 2\right)\right] = 3$
apply the definition of log (natural, with base $e$) to get:
$x \left(x - 2\right) = {e}^{3}$
rearranging:
${x}^{2} - 2 x - {e}^{3} = 0$

Use the Quadratic Formula to solve for $x$ as:
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 4 {e}^{3}}}{2}$
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 \left(1 + {e}^{3}\right)}}{2}$
${x}_{1 , 2} = \frac{2 \pm 2 \sqrt{1 + {e}^{3}}}{2}$
${x}_{1 , 2} = 1 \pm \sqrt{1 + {e}^{3}}$
Now check whether the two solutions are allowed or not by substituting them into the original equation (you'll find that only one works).