与えられた数学の問題を解く。問題は、指数の計算、式の計算、方程式、不等式に関するものである。

1. 問題の内容

与えられた数学の問題を解く。問題は、指数の計算、式の計算、方程式、不等式に関するものである。

2. 解き方の手順

**

1. 指数の計算**

(1) $(-3)^0 = 1$ (どんな数も0乗すると1になる)

(2) $10^{-1} = \frac{1}{10}$ (負の指数は逆数を意味する)

(3) $7^{-2} = \frac{1}{7^2} = \frac{1}{49}$

(4) $(-6)^{-3} = \frac{1}{(-6)^3} = \frac{1}{-216} = -\frac{1}{216}$

**

2. 式の計算**

(1) $49^{\frac{1}{2}} = \sqrt{49} = 7$

(2) $8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16$

(3) $16^{-\frac{3}{4}} = \frac{1}{16^{\frac{3}{4}}} = \frac{1}{(16^{\frac{1}{4}})^3} = \frac{1}{2^3} = \frac{1}{8}$

(4) $100^{-\frac{3}{2}} = \frac{1}{100^{\frac{3}{2}}} = \frac{1}{(100^{\frac{1}{2}})^3} = \frac{1}{10^3} = \frac{1}{1000}$

**

3. 式の計算**

(1) $2^{\frac{5}{6}} \times 2^{-\frac{1}{2}} \div 2^{\frac{2}{3}} = 2^{\frac{5}{6} - \frac{1}{2} - \frac{2}{3}} = 2^{\frac{5}{6} - \frac{3}{6} - \frac{4}{6}} = 2^{-\frac{2}{6}} = 2^{-\frac{1}{3}}$

(2) $27^{\frac{2}{3}} \times 3^{-\frac{1}{6}} \times 81^{-\frac{1}{12}} = (3^3)^{\frac{2}{3}} \times 3^{-\frac{1}{6}} \times (3^4)^{-\frac{1}{12}} = 3^2 \times 3^{-\frac{1}{6}} \times 3^{-\frac{1}{3}} = 3^{2 - \frac{1}{6} - \frac{1}{3}} = 3^{2 - \frac{1}{6} - \frac{2}{6}} = 3^{2 - \frac{3}{6}} = 3^{2 - \frac{1}{2}} = 3^{\frac{3}{2}}$

(3) $\sqrt[4]{25} \times \sqrt[12]{25} \div \sqrt[3]{25} = (25)^{\frac{1}{4}} \times (25)^{\frac{1}{12}} \div (25)^{\frac{1}{3}} = (5^2)^{\frac{1}{4}} \times (5^2)^{\frac{1}{12}} \div (5^2)^{\frac{1}{3}} = 5^{\frac{1}{2}} \times 5^{\frac{1}{6}} \div 5^{\frac{2}{3}} = 5^{\frac{1}{2} + \frac{1}{6} - \frac{2}{3}} = 5^{\frac{3}{6} + \frac{1}{6} - \frac{4}{6}} = 5^0 = 1$

(4) $\sqrt{8} \times \sqrt[6]{32} \times \sqrt[3]{4} = (2^3)^{\frac{1}{2}} \times (2^5)^{\frac{1}{6}} \times (2^2)^{\frac{1}{3}} = 2^{\frac{3}{2}} \times 2^{\frac{5}{6}} \times 2^{\frac{2}{3}} = 2^{\frac{9}{6} + \frac{5}{6} + \frac{4}{6}} = 2^{\frac{18}{6}} = 2^3 = 8$

**

4. 方程式**

(1) $(\frac{1}{3})^x = \frac{1}{81}$

$(\frac{1}{3})^x = (\frac{1}{3})^4$

$x = 4$

(2) $32^x = 8$

$(2^5)^x = 2^3$

$2^{5x} = 2^3$

$5x = 3$

$x = \frac{3}{5}$

(3) $16^{1-x} = (\frac{1}{64})^{x+1}$

$(2^4)^{1-x} = (2^{-6})^{x+1}$

$2^{4-4x} = 2^{-6x-6}$

$4-4x = -6x-6$

$2x = -10$

$x = -5$

**

5. 不等式**

(1) $2^x < 128$

$2^x < 2^7$

$x < 7$

(2) $3^x > \frac{1}{27}$

$3^x > 3^{-3}$

$x > -3$

(3) $(\frac{1}{9})^x > 27$

$(3^{-2})^x > 3^3$

$3^{-2x} > 3^3$

$-2x > 3$

$x < -\frac{3}{2}$

(4) $5^{x-2} \le 125$

$5^{x-2} \le 5^3$

$x-2 \le 3$

$x \le 5$

(5) $3^{2-x} > 9^x$

$3^{2-x} > (3^2)^x$

$3^{2-x} > 3^{2x}$

$2-x > 2x$

$2 > 3x$

$x < \frac{2}{3}$

(6) $(\frac{1}{8})^x \ge (\frac{1}{2})^{x+1}$

$(2^{-3})^x \ge (2^{-1})^{x+1}$

$2^{-3x} \ge 2^{-x-1}$

$-3x \ge -x-1$

$-2x \ge -1$

$x \le \frac{1}{2}$

3. 最終的な答え

1. (1) 1 (2) 1/10 (3) 1/49 (4) -1/216

2. (1) 7 (2) 16 (3) 1/8 (4) 1/1000

3. (1) $2^{-\frac{1}{3}}$ (2) $3^{\frac{3}{2}}$ (3) 1 (4) 8

4. (1) 4 (2) 3/5 (3) -5

5. (1) $x<7$ (2) $x>-3$ (3) $x<-\frac{3}{2}$ (4) $x \le 5$ (5) $x<\frac{2}{3}$ (6) $x \le \frac{1}{2}$