# How do you solve log (x^2-9) = log (5x+5)?

Aug 1, 2015

$\textcolor{red}{x = 7}$

#### Explanation:

$\log \left({x}^{2} - 9\right) = \log \left(5 x + 5\right)$

Convert the logarithmic equation to an exponential equation.

${10}^{\log \left({x}^{2} - 9\right)} = {10}^{\log \left(5 x + 5\right)}$

Remember that ${10}^{\log} x = x$, so

${x}^{2} - 9 = 5 x + 5$

Move all terms to the left hand side.

${x}^{2} - 9 - 5 x - 5 = 0$

Combine like terms.

${x}^{2} - 5 x - 14 = 0$

Factor.

$\left(x - 7\right) \left(x + 2\right) = 0$

$x - 7 = 0$ and $x + 2 = 0$

$x = 7$ and $x = - 2$

Check:

$\log \left({x}^{2} - 9\right) = \log \left(5 x + 5\right)$

If $x = 7$

$\log \left({7}^{2} - 9\right) = \log \left(5 \left(7\right) + 5\right)$

$\log \left(49 - 9\right) = \log \left(35 + 5\right)$

$\log 40 = \log 40$

$x = 7$ is a solution.

If $x = - 2$,

$\log \left({\left(- 2\right)}^{2} - 9\right) = \log \left(5 \left(- 2\right) + 5\right)$

$\log \left(4 - 9\right) = \log \left(- 10 + 5\right)$

$\log \left(- 5\right) = \log \left(- 5\right)$

#log(-5) is not defined,

$x = - 2$ is a spurious solution.