배터리 충전 지수식

[출제의도] 지수함수를 이용하여 실생활 문제를 해결한다. $a > 0$에서 $\displaystyle 0 < 2 ^{- \frac{2}{a}} < 1$이므로 $\displaystyle 1 - 2 ^{- \frac{2}{a}} > 0$이다. $\displaystyle \frac{Q ( 4 )}{Q ( 2 )}$$\displaystyle = \frac{Q _{0} \left( 1 - 2 ^{- \frac{4}{a}} \right)}{Q _{0} \left( 1 - 2 ^{- \frac{2}{a}} \right)}$$\displaystyle = \frac{1 - \left( 2 ^{- \frac{2}{a}} \right) ^{2}}{1 - 2 ^{- \frac{2}{a}}}$$\displaystyle = \frac{\left( 1 - 2 ^{- \frac{2}{a}} \right) \left( 1 + 2 ^{- \frac{2}{a}} \right)}{1 - 2 ^{- \frac{2}{a}}}$$\displaystyle = 1 + 2 ^{- \frac{2}{a}}$ $\displaystyle \frac{Q ( 4 )}{Q ( 2 )}$$\displaystyle = \frac{3}{2}$에서 $\displaystyle 1 + 2 ^{- \frac{2}{a}} = \frac{3}{2}$ $\displaystyle 2 ^{- \frac{2}{a}} = \frac{1}{2} = 2 ^{- 1}$ $\displaystyle - \frac{2}{a} = \mathit{-} 1$에서 $a = 2$ [다른 풀이] $\displaystyle \frac{Q ( 4 )}{Q ( 2 )}$$\displaystyle = \frac{3}{2}$에서 $2 Q ( 4 ) = 3 Q ( 2 )$ $\displaystyle 2 Q _{0} \left( 1 - 2 ^{- \frac{4}{a}} \right) = 3 Q _{0} \left( 1 - 2 ^{- \frac{2}{a}} \right)$ $\displaystyle 2 ^{- \frac{2}{a}} = t$로 놓으면 $a > 0$이므로 $0 < t < 1$이다. $2 ( 1 - t ^{2} ) = 3 \left( 1 - t \right)$ $2 ( 1 - t ) ( 1 + t ) = 3 ( 1 - t )$ $2 ( 1 + t ) = 3$ $\displaystyle t = \frac{1}{2}$ 즉 $\displaystyle 2 ^{- \frac{2}{a}} = 2 ^{- 1}$ $\displaystyle - \frac{2}{a} = \mathit{-} 1$에서 $a = 2$