지표와 가수

[출제의도] 지표와 가수를 활용하여 문제해결하기 정수 $k$($k \geq 0$)에 대하여 $10 ^{k} \leq x < 10 ^{k + 1}$에서 $\log x$의 지표와 가수는 $f ( x ) = k$, $g ( x ) = \log x - k$이므로 $y = \left\{ f ( x ) + 1 \right\} g \left( x \right) = \left( {k + 1} \right) \left( \log x - k \right)$가 $y = n$과 만나는 점의 $x$좌표는 $\left( {k + 1} \right) \left( \log x - k \right) = n$ $\displaystyle \log x = k + \frac{n}{k + 1}$ (단, $n < k + 1$) $\therefore$ $\displaystyle x = 10 ^{k + \frac{n}{k + 1}}$ $n = 1$일 때, $\displaystyle x = 10 ^{1 + \frac{1}{2}}$, $\displaystyle 10 ^{2 + \frac{1}{3}}$, $\displaystyle 10 ^{3 + \frac{1}{4}}$, $\cdots$ $\therefore$ $\displaystyle a _{1} = 10 ^{1 + \frac{1}{2}}$ $n = 2$일 때, $\displaystyle x = 10 ^{2 + \frac{2}{3}}$, $\displaystyle 10 ^{3 + \frac{2}{4}}$, $\displaystyle 10 ^{4 + \frac{2}{5}}$, $\cdots$ $\therefore$ $\displaystyle a _{2} = 10 ^{2 + \frac{2}{3}}$ $n = 3$일 때, $\displaystyle x = 10 ^{3 + \frac{3}{4}}$, $\displaystyle 10 ^{4 + \frac{3}{5}}$, $\displaystyle 10 ^{5 + \frac{3}{6}}$, $\cdots$ $\therefore$ $\displaystyle a _{3} = 10 ^{3 + \frac{3}{4}}$ $\vdots$ $\therefore$ $\displaystyle a _{n} = 10 ^{n + \frac{n}{n + 1}}$ $\displaystyle \sum\limits _{n = 1} ^{10} \left( \log _{} a _{n} + \frac{1}{n + 1} \right) = \sum\limits _{n = 1} ^{10} \left( \log _{} 10 ^{n + \frac{n}{n + 1}} + \frac{1}{n + 1} \right)$ $\displaystyle = \sum\limits _{n = 1} ^{10} \left( n + 1 \right) = 65$

$y = \left\{ f ( x ) + 1 \right\} g \left( x \right)$