로그 조건으로 값 구하기

$\log _{a} b = 3$에서 $b = a ^{3}$ 이를 $\displaystyle \log _{3} \frac{b}{a} = \frac{1}{2}$에 대입하면 $\displaystyle \log _{3} \frac{a ^{3}}{a} = \frac{1}{2}$, $\displaystyle \log _{3} a ^{2} = \frac{1}{2}$ $\displaystyle 2 \log _{3} a = \frac{1}{2}$, $\displaystyle \log _{3} a = \frac{1}{4}$ $\therefore$ $\displaystyle a = 3 ^{\frac{1}{4}}$, $\displaystyle b = 3 ^{\frac{3}{4}}$ 따라서 $\displaystyle ab = 3 ^{\frac{1}{4}} \times 3 ^{\frac{3}{4}} = 3 ^{\frac{1}{4} + \frac{3}{4}} = 3 ^{1} = 3$이므로 $\displaystyle \log _{9} ab = \log _{9} 3 = \log _{3 ^{2}} 3 = \frac{1}{2} \log _{3} 3 = \frac{1}{2}$ [다른 풀이] $\log _{a} b = 3$에서 $\displaystyle \frac{\log _{3} {b}}{\log _{3} {a}} = 3$, $\log _{3} {b} = 3 \log _{3} {a}$ $\cdots$$\cdots$ ㉠ $\displaystyle \log _{3} \frac{b}{a} = \frac{1}{2}$에서 $\displaystyle \log _{3} {b} - \log _{3} {a} = \frac{1}{2}$ $\cdots$$\cdots$ ㉡ ㉠, ㉡에서 $\log _{3} {a} = x$, $\log _{3} {b} = y$라 하면 $y = 3 x$, $\displaystyle y - x = \frac{1}{2}$ 두 식을 연립하여 풀면 $\displaystyle x = \frac{1}{4}$, $\displaystyle y = \frac{3}{4}$ $\therefore$ $\log _{9} {ab}$$= \log _{3 ^{2}} ab$ $\displaystyle = \frac{1}{2} \log _{3} ab$ $\displaystyle = \frac{1}{2} \left( \log _{3} {a} + \log _{3} {b} \right)$ $\displaystyle = \frac{1}{2} \left( x + y \right)$ $\displaystyle = \frac{1}{2} \left( \frac{1}{4} + \frac{3}{4} \right)$ $\displaystyle = \frac{1}{2}$