# How do you solve log (2x + 1) - log (x - 2) = 1?

Mar 16, 2016

$x = \frac{21}{8}$

#### Explanation:

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

$\log \left(2 x + 1\right) - \log \left(x - 2\right) = 1$

$\log \left(\frac{2 x + 1}{x - 2}\right) = 1$

$2$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{p u r p \le}{b}}^{\textcolor{\mathmr{and} a n \ge}{x}}\right) = \textcolor{\mathmr{and} a n \ge}{x}$, to rewrite the right side of the equation.

$\log \left(\frac{2 x + 1}{x - 2}\right) = \log \left(10\right)$

$3$. Since the equation now follows a "$\log = \log$" situation, where the bases are the same on both sides of the equation, rewrite the equation without the "log" portion.

$\frac{2 x + 1}{x - 2} = 10$

$4$. Solve for $x$.

$2 x + 1 = 10 \left(x - 2\right)$

$2 x + 1 = 10 x - 20$

$8 x = 21$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{21}{8} \textcolor{w h i t e}{\frac{a}{a}} |}}}$