# How do you solve Log3x=log4+log(x+3)?

Apr 19, 2016

x > 0. The negative solution $x = - 12$ is inadmissible for the given equation. So there is no real solution.

#### Explanation:

x > 0.
Use $\log a - \log b = \log \left(\frac{a}{b}\right)$

$\log \left(\frac{3 x}{x + 3}\right) = \log 4$

Som $\frac{3 x}{x + 3} = 4$

$x = - 12$.

As x > 0 for log x, $x = - 12$ is inadmissible.