# How do you solve Log (x+5) - log (x-1) = log (x+2) - log (x-3)?

Nov 21, 2015

$x = 13$

#### Explanation:

To do this problem, you must know that $\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$ and that $\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$.

$\log \left(x + 5\right) - \log \left(x - 1\right) = \log \left(x + 2\right) - \log \left(x - 3\right)$

$\log \left(x + 5\right) + \log \left(x - 3\right) - \log \left(x - 1\right) - \log \left(x + 2\right) = 0$

$\log \left(\frac{\left(x + 5\right) \left(x - 3\right)}{\left(x - 1\right) \left(x + 2\right)}\right) = 0$

Remember that $\log \left(a\right)$ is another way of writing ${\log}_{10} \left(a\right)$.

${10}^{\log \left(\frac{\left(x + 5\right) \left(x - 3\right)}{\left(x - 1\right) \left(x + 2\right)}\right)} = {10}^{0}$

$\frac{\left(x + 5\right) \left(x - 3\right)}{\left(x - 1\right) \left(x + 2\right)} = 1$

$\left(x + 5\right) \left(x - 3\right) = \left(x - 1\right) \left(x + 2\right)$

${x}^{2} + 2 x - 15 = {x}^{2} + x - 2$

$2 x - 15 = x - 2$

$x = 13$