지수와 로그 조건

[출제의도] 지수법칙 이해하기 $2 ^{a} = 3 ^{b} = k$$\left( \right.$$k > 1$$)$이라 놓으면 $\displaystyle 2 = k ^{\frac{1}{a}}$, $\displaystyle 3 = k ^{\frac{1}{b}}$ $( a - 2 ) ( b - 2 ) = 4$에서 $\displaystyle \frac{a + b}{ab} = \frac{1}{2}$ $\displaystyle k ^{\frac{1}{a}} \times k ^{\frac{1}{b}} = k ^{\frac{a + b}{ab}} = k ^{{\frac{1}{2}}} = 6$ 따라서 $\displaystyle 4 ^{a} \times 3 ^{- b} = \frac{\left( 2 ^{a} \right) ^{2}}{3 ^{b}} = \frac{k ^{2}}{k} = k = 36$ (별해) $2 ^{a} = 3 ^{b} = k$$\left( \right.$$k > 1$$)$이라 놓으면 $a = \log _{2} k$, $b = \log _{3} k$ $( a - 2 ) ( b - 2 ) = ( \log _{2} k - 2 ) ( \log _{3} k - 2 )$ $= \log _{2} k \times \log _{3} k - 2 \log _{2} k - 2 \log _{3} k + 4 = 4$ $\log _{2} k \times \log _{3} k - 2 ( \log _{2} k + \log _{3} k ) = 0$ $\displaystyle \frac{\log _{2} k + \log _{3} k}{\log _{2} k \times \log _{3} k} = \frac{1}{2}$ $\displaystyle \frac{1}{\log _{2} k} + \frac{1}{\log _{3} k} = \frac{1}{2}$ $\displaystyle \log _{k} 2 + \log _{k} 3 = \log _{k} 6 = \frac{1}{2}$, 즉 $\displaystyle k ^{\frac{1}{2}} = 6$ 따라서 $\displaystyle 4 ^{a} \times 3 ^{- b} = \frac{\left( 2 ^{a} \right) ^{2}}{3 ^{b}} = \frac{k ^{2}}{k} = k = 36$