与えられた各組の数について、それらの大小関係を不等号を用いて表す問題です。 (1) $2^{\frac{3}{2}}$, $2^{-2}$, $2^5$, $1$ (2) $0.7^5$, $0.7^{-\frac{1}{2}}$, $1$, $0.7^2$ (3) $\sqrt[5]{8}$, $\sqrt[6]{16}$, $\sqrt[8]{64}$ (4) $(\frac{1}{3})^{-2}$, $\frac{1}{3}$, $(\frac{1}{27})^{\frac{1}{5}}$, $\sqrt[7]{\frac{1}{81}}$

1. 問題の内容

与えられた各組の数について、それらの大小関係を不等号を用いて表す問題です。

(1) $2^{\frac{3}{2}}$, $2^{-2}$, $2^5$, $1$

(2) $0.7^5$, $0.7^{-\frac{1}{2}}$, $1$, $0.7^2$

(3) $\sqrt[5]{8}$, $\sqrt[6]{16}$, $\sqrt[8]{64}$

(4) $(\frac{1}{3})^{-2}$, $\frac{1}{3}$, $(\frac{1}{27})^{\frac{1}{5}}$, $\sqrt[7]{\frac{1}{81}}$

2. 解き方の手順

(1)

$2^{\frac{3}{2}}=\sqrt{2^3} = \sqrt{8} \approx 2.83$

$2^{-2} = \frac{1}{2^2} = \frac{1}{4} = 0.25$

$2^5 = 32$

$1 = 1$

したがって、$2^{-2} < 1 < 2^{\frac{3}{2}} < 2^5$

(2)

$0.7^5 \approx 0.168$

$0.7^{-\frac{1}{2}} = (\frac{7}{10})^{-\frac{1}{2}} = (\frac{10}{7})^{\frac{1}{2}} = \sqrt{\frac{10}{7}} \approx \sqrt{1.43} \approx 1.19$

$1 = 1$

$0.7^2 = 0.49$

したがって、$0.7^5 < 0.7^2 < 1 < 0.7^{-\frac{1}{2}}$

(3)

$\sqrt[5]{8} = 8^{\frac{1}{5}} = (2^3)^{\frac{1}{5}} = 2^{\frac{3}{5}} = 2^{0.6}$

$\sqrt[6]{16} = 16^{\frac{1}{6}} = (2^4)^{\frac{1}{6}} = 2^{\frac{4}{6}} = 2^{\frac{2}{3}} = 2^{0.666...}$

$\sqrt[8]{64} = 64^{\frac{1}{8}} = (2^6)^{\frac{1}{8}} = 2^{\frac{6}{8}} = 2^{\frac{3}{4}} = 2^{0.75}$

指数関数$y = 2^x$は単調増加なので、$0.6 < 0.666... < 0.75$より、$2^{0.6} < 2^{0.666...} < 2^{0.75}$

したがって、$\sqrt[5]{8} < \sqrt[6]{16} < \sqrt[8]{64}$

(4)

$(\frac{1}{3})^{-2} = 3^2 = 9$

$\frac{1}{3} \approx 0.33$

$(\frac{1}{27})^{\frac{1}{5}} = (3^{-3})^{\frac{1}{5}} = 3^{-\frac{3}{5}} = \frac{1}{3^{\frac{3}{5}}} = \frac{1}{\sqrt[5]{3^3}} = \frac{1}{\sqrt[5]{27}}$

$1 < \sqrt[5]{27} < 2$なので、$\frac{1}{2} < \frac{1}{\sqrt[5]{27}} < 1$

$\sqrt[7]{\frac{1}{81}} = (\frac{1}{81})^{\frac{1}{7}} = (3^{-4})^{\frac{1}{7}} = 3^{-\frac{4}{7}} = \frac{1}{3^{\frac{4}{7}}} = \frac{1}{\sqrt[7]{3^4}} = \frac{1}{\sqrt[7]{81}}$

$1 < \sqrt[7]{81} < 2$なので、$\frac{1}{2} < \frac{1}{\sqrt[7]{81}} < 1$

$3^{\frac{3}{5}} = 3^{0.6} \approx 1.93$

$3^{\frac{4}{7}} \approx 3^{0.57} \approx 1.83$

よって、$\frac{1}{1.93} \approx 0.52$, $\frac{1}{1.83} \approx 0.55$

したがって、$(\frac{1}{27})^{\frac{1}{5}} < \sqrt[7]{\frac{1}{81}} < \frac{1}{3} < (\frac{1}{3})^{-2}$

3. 最終的な答え

(1) $2^{-2} < 1 < 2^{\frac{3}{2}} < 2^5$

(2) $0.7^5 < 0.7^2 < 1 < 0.7^{-\frac{1}{2}}$

(3) $\sqrt[5]{8} < \sqrt[6]{16} < \sqrt[8]{64}$

(4) $(\frac{1}{27})^{\frac{1}{5}} < \sqrt[7]{\frac{1}{81}} < \frac{1}{3} < (\frac{1}{3})^{-2}$