# How do you solve lnx+ln(x-2)=1?

Jul 20, 2016

I found: $x = 1 + \sqrt{1 + e}$

#### Explanation:

We can use the property of logs that allows us to change the sum into a product of arguments as:
$\ln \left[x \cdot \left(x - 2\right)\right] = 1$
then use the definition of log, applied to our natural log, to write:
$x \left(x - 2\right) = {e}^{1}$
${x}^{2} - 2 x - e = 0$
solve for $x$ to get:
${x}_{1 , 2} = \frac{2 \pm \sqrt{4 + 4 e}}{2} = \frac{2 \pm 2 \sqrt{1 + e}}{2}$
${x}_{1} = 1 + \sqrt{1 + e}$
${x}_{2} = 1 - \sqrt{1 + e}$ NOT (because negative)

Jul 20, 2016

Use the properties:

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

$\ln x = y \implies x = {e}^{y}$

${x}_{1} = 2.928$

${x}_{2} = - 0.928$

#### Explanation:

$\ln x + \ln \left(x - 2\right) = 1$

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

$\ln \left(x \cdot \left(x - 2\right)\right) = 1$

To add an $\ln$ function to the right side we use $x = \ln {e}^{x}$

$\ln \left(x \cdot \left(x - 2\right)\right) = \ln {e}^{1}$

$\ln x$ is a $1 - 1$ function:

$x \cdot \left(x - 2\right) = e$

${x}^{2} - 2 x - e = 0$

Δ=(-2)^2-4*1*(-e)=4+4e
${x}_{1 , 2} = \frac{- \left(- 2\right) \pm \sqrt{4 + 4 e}}{2 \cdot 1}$
From the calculator $\sqrt{4 + 4 e} = 3.857$
${x}_{1} = 2.928$
${x}_{2} = - 0.928$