로그방정식

$\log _{3} \left( x - 4 \right) = \log _{3 ^{2}} \left( 5 x + 4 \right)$ $\displaystyle \log _{3} \left( x - 4 \right) = \frac{1}{2} \log _{3} \left( 5 x + 4 \right)$ $2 \log _{3} \left( x - 4 \right) = \log _{3} \left( 5 x + 4 \right)$ $\log _{3} \left( x - 4 \right) ^{2} = \log _{3} \left( 5 x + 4 \right)$ $\left( x - 4 \right) ^{2} = 5 x + 4$, $x ^{2} - 13 x + 12 = 0$ $\left( x - 1 \right) \left( x - 12 \right) = 0$ $\therefore x = 1$ 또는 $x = 12$ 이때 진수조건에 의하여 $x > 4$이므로 $\alpha = 12$