対数不等式の問題

■問題

$3$\displaystyle{ \hspace{1}{ \log }_{3}9\cdot \hspace{1}{ \log }_{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }4\leq \hspace{1}{ \log }_{9}x\leq \hspace{1}{ \log }_{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }4+\hspace{1}{ \log }_{3}2 }$

■答

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 256 \hspace{2}}\leq x\leq \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$

■解説

$3$\displaystyle{\hspace{1}{\log }_{3}9\hspace{1}{\log }_{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }4}$ $3$\displaystyle{\leq \hspace{1}{\log }_{9}x}$ $3$\displaystyle{\leq \hspace{1}{\log }_{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }4+\hspace{1}{\log }_{3}2}$

底の変換公式を用いて，底を $3$3$ にそろえる．その後，対数計算の基本を利用して式を簡単にする．

ここで

$3$\displaystyle{ \hspace{1}{ \log }_{3} 9}$ $3$\displaystyle{=\hspace{1}{\log }_{3}{3}^{2}}$ $3$\displaystyle{ =2\hspace{1}{ \log }_{3}3 }$ $3$\displaystyle{=2}$

$3$\displaystyle{ \hspace{1}{ \log }_{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }4 }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{ \log }_{3}4 \hspace{2}}{\hspace{2} \hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} \hspace{2}} }$ $3$\displaystyle{ =- \hspace{1}{ \log }_{3}4 }$ $3$\displaystyle{ =\hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$

$3$\displaystyle{\hspace{1}{\log }_{9}x}$ $3$\displaystyle{=\frac{\hspace{2} \hspace{1}{ \log }_{3}x \hspace{2}}{\hspace{2} \hspace{1}{ \log }_{3}9 \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{\log }_{3}x}$

よって

$3$\displaystyle{2\frac{\hspace{2} \hspace{1}{ \log }_{3}4 \hspace{2}}{\hspace{2} \hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} \hspace{2}}}$ $3$\displaystyle{\leq \frac{\hspace{2} \hspace{1}{ \log }_{3}x \hspace{2}}{\hspace{2} \hspace{1}{ \log }_{3}9 \hspace{2}}}$ $3$\displaystyle{\leq \frac{\hspace{2} \hspace{1}{ \log }_{3}4 \hspace{2}}{\hspace{2} \hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} \hspace{2}}+\hspace{1}{\log }_{3}2}$

$3$\displaystyle{2 $ - \hspace{1}{ \log }_{3}4 $ }$ $3$\displaystyle{\leq \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{\log }_{3}x}$ $3$\displaystyle{\leq - \hspace{1}{\log }_{3}4+\hspace{1}{\log }_{3}2}$

さらに，ここで

$3$\displaystyle{2 $ - \hspace{1}{ \log }_{3}4 $ }$ $3$\displaystyle{= 2\hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$ $3$\displaystyle{ =\hspace{1}{ \log }_{3}{ $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} $ }^{2} }$ $3$\displaystyle{ =\hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 16 \hspace{2}} }$

$3$\displaystyle{ - \hspace{1}{ \log }_{3}4+\hspace{1}{ \log }_{3}2 }$ $3$\displaystyle{=\hspace{1}{\log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}+\hspace{1}{\log }_{3}2}$ $3$\displaystyle{=\hspace{1}{\log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{\log }_{3}x}$ $3$\displaystyle{=\hspace{1}{\log }_{3}{x}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }}$ $3$\displaystyle{=\hspace{1}{\log }_{3}\sqrt{x}}$

よって

$3$\displaystyle{ 2\hspace{1}{ \log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}} }$ $3$\displaystyle{\leq \hspace{1}{\log }_{3}{x}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }}$ $3$\displaystyle{\leq \hspace{1}{\log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}+\hspace{1}{\log }_{3}2}$

$3$\displaystyle{\hspace{1}{\log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 16 \hspace{2}}}$ $3$\displaystyle{≦\hspace{1}{\log }_{3}\sqrt{x}}$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
$3$\displaystyle{\leq \hspace{1}{\log }_{3}\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}}$

底が $3$3$ で $3$1$ より大きい．よって，真数の大小関係は以下のようになる．

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} 16 \hspace{2}}}$ $3$\displaystyle{\leq \sqrt{x}}$ $3$\displaystyle{\leq \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} 256 \hspace{2}}}$ $3$\displaystyle{\leq x}$ $3$\displaystyle{\leq \frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}}$

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作成：学生スタッフ

最終更新日： 2023年11月28日