로그 연립 등식

$a ^{2} b ^{3} = 64$ -----① $\left( \log _{a} c \right) ^{2} - 2 \left( \log _{b} c \right) ^{2} = - \left( \log _{a} c \right) \left( \log _{b} c \right)$ ----- ② ① $\leftrightarrow$ $2 \log _{2} a + 3 \log _{2} b = 6$ -----$\text{ ①}'$ ② $\leftrightarrow$ $\displaystyle 3 {\frac{\log _{a} c}{\log _{b} c}} - 2 {\frac{\log _{b} c}{\log _{a} c}} = - 1$ $\leftrightarrow$ $\displaystyle 3 {\frac{\log _{c} b}{\log _{c} a}} - 2 {\frac{\log _{c} a}{\log _{c} b}} = - 1$ $\leftrightarrow$ $\displaystyle 3 {\frac{\log _{2} b}{\log _{2} a}} - 2 {\frac{\log _{2} a}{\log _{2} b}} = - 1$ $\leftrightarrow$ $3 ( \log _{2} b ) ^{2} - 2 ( \log _{2} a ) ^{2} = - \log _{2} a \log _{2} b$ $\leftrightarrow$ $\displaystyle 3 ( \log _{2} b ) ^{2} - 2 \left\{ 3 - \frac{3}{2} \log _{2} b \right\} ^{2}$ $\displaystyle = - \left\{ 3 - \frac{3}{2} \log _{2} b \right\} \log _{2} b$ ($\because \text{ ①}'$) -------$\text{ ②}'$ $\log _{2} b = x$ 라고 하면 $\text{ ②}'$$\leftrightarrow$$3 x ^{2} - 2 ( 3 - \frac{3}{2} x ) ^{2} = - 3 x + \frac{3}{2} x ^{2}$ $\leftrightarrow$ $x ^{2} - 7 x + 6 = 0$ $\therefore$ $x = 1 or 6$ [i] $x = 1$ 이면 (즉, $\log _{2} b = 1$) $\displaystyle \log _{2} a = 3 - \frac{3}{2} \log _{2} b = \frac{3}{2}$ ($\because \text{ ①}'$) [ii] $x = 6$ 이면 (즉, $\log _{2} b = 6$) $\displaystyle \log _{2} a = 3 - \frac{3}{2} \log _{2} b = - 6$ ($\because \text{ ①}'$) [문제의 조건을 만족하지 않음] ($\because$ 문제의 조건에서 $a , b , c$ 가 1보다 크기 때문에 $\log _{2} a , \log _{2} b , \log _{2} c$ 는 0보다 크다. ) $\therefore$ $\displaystyle \log _{2} a = \frac{3}{2}$, $\log _{2} b = 1$ $\therefore$ $\displaystyle \log _{2} ab = \log _{2} a + \log _{2} b = \frac{3}{2} + 1 = \frac{5}{2}$