How do you solve log(3x + 7)+ log(x - 2)= 1?

Mar 16, 2016

$x = \frac{8}{3}$

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 left side of the equation.

$\log \left(3 x + 7\right) + \log \left(x - 2\right) = 1$

$\log \left(\left(3 x + 7\right) \left(x - 2\right)\right) = 1$

$\log \left(3 {x}^{2} + x - 14\right) = 1$

$2$. 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.

$\log \left(3 {x}^{2} + x - 14\right) = \log \left(10\right)$

$3$. 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.

$3 {x}^{2} + x - 14 = 10$

$4$. Subtract $10$ from both sides of the equation.

$3 {x}^{2} + x - 24 = 0$

$5$. Factor the quadratic equation.

$\left(3 x - 8\right) \left(x + 3\right) = 0$

$6$. Set each factor to $0$ and solve for $x$.

$3 x - 8 = 0 \textcolor{w h i t e}{X X X X X X X} x + 3 = 0$

$3 x = 8 \textcolor{w h i t e}{X X X X X X X X X} \textcolor{red}{\cancel{\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = - 3 \textcolor{w h i t e}{\frac{a}{a}} |}}}}}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{8}{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$7$. However, $\textcolor{red}{x = - 3}$ does not satisfy the equation since it produces a log with a negative number, and you $\textcolor{t e a l}{\text{can't log a negative number}}$.

For example, when you substitute $\textcolor{red}{x = - 3}$ back into the original equation:

$\log \left(3 x + 7\right) + \log \left(x - 2\right) = 1$

$\log \left(3 \textcolor{red}{\left(- 3\right)} + 7\right) + \log \left(\textcolor{red}{\left(- 3\right)} - 2\right) = 1$

$\log \left(- 9 + 7\right) + \log \left(- 5\right) = 1$

$\textcolor{t e a l}{\log \left(- 2\right)} + \textcolor{t e a l}{\log \left(- 5\right)} = 1$

For this reason, $\textcolor{red}{x = - 3}$ is not a solution of the equation.

$\therefore$, $x = \frac{8}{3}$.