与えられた指数方程式・不等式を解く。 (1) $4^x = 64$ (2) $25^x = \frac{1}{125}$ (3) $(\frac{1}{8})^x = 16$ (4) $(\frac{1}{9})^{3x-1} = (\frac{1}{27})^x$ (5) $2^x > 64$ (6) $81^x \le \frac{1}{243}$ (7) $(\frac{1}{4})^x < \frac{1}{8}$ (8) $(\frac{1}{36})^{2x-1} \ge 6^{x+3}$

1. 問題の内容

与えられた指数方程式・不等式を解く。

(1) $4^x = 64$

(2) $25^x = \frac{1}{125}$

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

(4) $(\frac{1}{9})^{3x-1} = (\frac{1}{27})^x$

(5) $2^x > 64$

(6) $81^x \le \frac{1}{243}$

(7) $(\frac{1}{4})^x < \frac{1}{8}$

(8) $(\frac{1}{36})^{2x-1} \ge 6^{x+3}$

2. 解き方の手順

(1)

$4^x = 64$

$4^x = 4^3$

$x = 3$

(2)

$25^x = \frac{1}{125}$

$(5^2)^x = 5^{-3}$

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

$2x = -3$

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

(3)

$(\frac{1}{8})^x = 16$

$(2^{-3})^x = 2^4$

$2^{-3x} = 2^4$

$-3x = 4$

$x = -\frac{4}{3}$

(4)

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

$(3^{-2})^{3x-1} = (3^{-3})^x$

$3^{-2(3x-1)} = 3^{-3x}$

$-2(3x-1) = -3x$

$-6x + 2 = -3x$

$2 = 3x$

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

(5)

$2^x > 64$

$2^x > 2^6$

$x > 6$

(6)

$81^x \le \frac{1}{243}$

$(3^4)^x \le 3^{-5}$

$3^{4x} \le 3^{-5}$

$4x \le -5$

$x \le -\frac{5}{4}$

(7)

$(\frac{1}{4})^x < \frac{1}{8}$

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

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

$-2x < -3$

$2x > 3$

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

(8)

$(\frac{1}{36})^{2x-1} \ge 6^{x+3}$

$(6^{-2})^{2x-1} \ge 6^{x+3}$

$6^{-2(2x-1)} \ge 6^{x+3}$

$-2(2x-1) \ge x+3$

$-4x + 2 \ge x+3$

$-1 \ge 5x$

$x \le -\frac{1}{5}$

3. 最終的な答え

(1) $x = 3$

(2) $x = -\frac{3}{2}$

(3) $x = -\frac{4}{3}$

(4) $x = \frac{2}{3}$

(5) $x > 6$

(6) $x \le -\frac{5}{4}$

(7) $x > \frac{3}{2}$

(8) $x \le -\frac{1}{5}$