# How do you solve Log(x+10) - log(x+4) = log x?

Dec 9, 2015

Apply properties of logarithms and solve the resulting quadratic equation to find
$x = 2$ or $x = - 5$

#### Explanation:

We will use the following properties:

• $\log \left(x\right) - \log \left(y\right) = \log \left(\frac{x}{y}\right)$

• ${e}^{\log} \left(x\right) = x$

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

$\implies \log \left(\frac{x + 10}{x + 4}\right) = \log \left(x\right)$

$\implies {e}^{\log} \left(\frac{x + 10}{x + 4}\right) = {e}^{\log} \left(x\right)$

$\implies \frac{x + 10}{x + 4} = x$

$\implies x + 10 = x \left(x + 4\right) = {x}^{2} + 4 x$

$\implies {x}^{2} + 3 x - 10 = 0$

$\implies \left(x + 5\right) \left(x - 2\right) = 0$

$\implies x = 2$ or $x = - 5$