# How do you solve ln x + ln (x+6) = 1/2 ln 9?

Jun 20, 2015

$x = - 3 + \setminus \sqrt{12}$
Use rules for adding logarithms and rule for bringing the coefficient inside the logarithm as an exponent.

#### Explanation:

$\ln \left(A\right) + \ln \left(B\right) = \ln \left(A B\right)$

so

$\ln \left(x\right) + \ln \left(x + 6\right) = \ln \left({x}^{2} + 6 x\right)$

$C \ln \left(D\right) = \ln \left({D}^{C}\right)$

so

$\frac{1}{2} \ln \left(9\right) = \ln \left({9}^{\frac{1}{2}}\right) = \ln \left(3\right)$

Making these substitutions,

$\ln \left(x\right) + \ln \left(x + 6\right) = \frac{1}{2} \ln \left(9\right)$

becomes

$\ln \left({x}^{2} + 6 x\right) = \ln \left(3\right)$

which requires,

${x}^{2} + 6 x = 3$

$\setminus \implies {x}^{2} + 6 x - 3 = 0$

$x = \frac{- 6 \setminus \pm \setminus \sqrt{{6}^{2} - 4 \left(1\right) \left(- 3\right)}}{2 \left(1\right)}$

$x = - 3 \setminus \pm \setminus \sqrt{12}$

$\ln \left(x\right)$ requires $x > 0$ so we choose the plus sign

$x = - 3 + \sqrt{12}$