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

##### 1 Answer
Nov 16, 2016

Combine the left into one term using the rule $\ln \left(a b\right) = \ln a + \ln b$

$\ln x + \ln \left(x + 3\right) = 1$
$\ln \left[\left(x\right) \left(x + 3\right)\right] = 1$
$\ln \left({x}^{2} + 3 x\right) = 1$

Rewrite using the definition of log:
${e}^{1} = {x}^{2} + 3 x$
$e = {x}^{2} + 3 x$

Now solve as a quadratic; first set one side equal to zero:
$0 = {x}^{2} + 3 x - e$

Use quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(3\right) \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(e\right)}}{2 \left(1\right)}$
$x = \frac{- 3 \pm \sqrt{9 - 4 e}}{2}$

$x \approx \frac{- 3 \pm \sqrt{- 1.87}}{2}$

Because we take the square root of a negative number, there are no solutions to the equation.