How do you solve ln(x-6)+ln(x+3)=ln22?

Aug 13, 2015

$\textcolor{red}{x = 8}$

Explanation:

$\ln \left(x - 6\right) + \ln \left(x + 3\right) = \ln 22$

Recall that $\ln a + \ln b = \ln a b$.

$\ln \left(x - 6\right) \left(x + 3\right) = \ln 22$

Convert the logarithmic equation to an exponential equation.

${e}^{\ln} \left(\left(x - 6\right) \left(x + 3\right)\right) = {e}^{\ln} 22$

Remember that ${e}^{\ln} x = x$, so

$\left(x - 6\right) \left(x + 3\right) = 22$

${x}^{2} - 3 x - 18 = 22$

${x}^{2} - 3 x - 40 = 0$

$\left(x - 8\right) \left(x + 5\right) = 0$

$x - 8 = 0$ and $x + 5 = 0$

$x = 8$ and $x = - 5$ are possible solutions.

Check:

$\ln \left(x - 6\right) + \ln \left(x + 3\right) = \ln 22$

If $x = 8$,

$\ln \left(8 - 6\right) + \ln \left(8 + 3\right) = \ln 22$

$\ln 2 + \ln 11 = \ln 22$

ln(2×11)=ln22

$\ln 22 = \ln 22$

$x = 8$ is a solution.

If $x = - 5$,

$\ln \left(- 5 - 6\right) + \ln \left(- 5 + 3\right) = \ln 22$

$\ln \left(- 11\right) + \ln \left(- 2\right) = \ln 22$

But $\ln \left(- 11\right)$ and $\ln \left(- 2\right)$ are not defined.

$x = - 5$ is not a solution.