로그방정식

방정식 $\displaystyle \log _{2} \left( x + 1 \right) - 5 = \log _{{\frac{1}{2}}} \left( x - 3 \right)$에서 진수의 조건에 의하여 $x > 3$ $\cdots$$\cdots$ ㉠ $\displaystyle \log _{2} \left( x + 1 \right) - 5 = \log _{{\frac{1}{2}}} \left( x - 3 \right)$ $\log _{2} \left( x + 1 \right) + \log _{2} \left( x - 3 \right) = 5$ $\log _{2} \left( x + 1 \right) \left( x - 3 \right) = \log _{2} 2 ^{5}$ $\left( x + 1 \right) \left( x - 3 \right) = 32$ $x ^{2} - 2 x - 35 = 0$, $\left( x - 7 \right) \left( x + 5 \right) = 0$ $\therefore$ $x = 7$ $\left( \because \text{㉠} \right)$