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

Sep 6, 2015

$x = 4$

#### Explanation:

First we simplify the left side by using the rule: $\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$

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

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

Then, we apply the rule: $\log \left(a\right) = b \iff {10}^{b} = a$

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

Then, we can apply the rule: ${10}^{\log} \left(k\right) = k$

$3 = \frac{x + 2}{x - 2}$

Now we can just solve the equation:

$3 x - 6 = x + 2$

$2 x = 8$

$x = 4$

Alternatively, we can skip a lot of steps by taking a shortcut:

Since $\log \left(\right)$ is a one-to-one function, $\log \left(a\right) = \log \left(b\right)$ means $a = b$.

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

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

$\frac{x + 2}{x - 2} = 3$

$x + 2 = 3 x - 6$

$8 = 2 x$

$4 = x$