How do you solve log (x+6)=1-log(x-5)?

Mar 20, 2016

$x = \frac{- 1 \pm \sqrt{161}}{2}$

Explanation:

$1$. Bring all the logs to the left side of the equation.

$\log \left(x + 6\right) = 1 - \log \left(x - 5\right)$

$\log \left(x + 6\right) + \log \left(x - 5\right) = 1$

$2$. 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 rewrite the left side of the equation.

$\log \left(\left(x + 6\right) \left(x - 5\right)\right) = 1$

$3$. 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.

$\log \left(\left(x + 6\right) \left(x - 5\right)\right) = \log \left(10\right)$

$4$. 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.

$\left(x + 6\right) \left(x - 5\right) = 10$

$5$. Expand the brackets.

${x}^{2} + x - 30 = 10$

$6$. Subtract $10$ from both sides.

${x}^{2} + x - 40 = 0$

$7$. Use the quadratic formula to solve for $x$.

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{a = 1} \textcolor{w h i t e}{X X X X X X} \textcolor{t e a l}{b = 1} \textcolor{w h i t e}{X X X X X X} \textcolor{v i o \le t}{c = - 40}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{t e a l}{1}\right) \pm \sqrt{{\left(\textcolor{t e a l}{1}\right)}^{2} - 4 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{1}\right) \left(\textcolor{v i o \le t}{- 40}\right)}}{2 \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{1}\right)}$

$x = \frac{- 1 \pm \sqrt{1 + 160}}{2}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{- 1 \pm \sqrt{161}}{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$