# How do you solve log 3x=log 4+log (x-3)?

Mar 17, 2016

$x = 12$

#### Explanation:

$1$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m} \cdot \textcolor{b l u e}{n}\right) = {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right) + {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{b l u e}{n}\right)$ to simplify the right side of the equation.

$\log \left(3 x\right) = \log 4 + \log \left(x - 3\right)$

$\log \left(3 x\right) = \log \left(4 \left(x - 3\right)\right)$

$\log \left(3 x\right) = \log \left(4 x - 12\right)$

$2$. Since the equation now follows a "$\log = \log$" situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.

$3 x = 4 x - 12$

$3$. Solve for $x$.

$3 x - 4 x = - 12$

$- x = - 12$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = 12 \textcolor{w h i t e}{\frac{a}{a}} |}}}$