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

Aug 15, 2015

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

#### Explanation:

If $a , b > 0$ then $\log \left(a\right)$ and $\log \left(b\right)$ are defined and

$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

So $\log \left(x\right) + \log \left(4\right) = \log \left(4 x\right)$

So our equation becomes:

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

Since $\log : \left(0 , \infty\right) \to \mathbb{R}$ is one-one, that means:

$x + 4 = 4 x$

Subtract $x$ from both sides to get $4 = 3 x$, then divide both sides by $3$ to get $x = \frac{4}{3}$.