次の方程式・不等式を解く問題です。 (1) $\sqrt[3]{9^x} = 3 \sqrt[4]{9^x}$ (2) $9^{x+1} + 80 \cdot 3^{x-1} - 1 = 0$ (3) $2^{2x} - 2^{x+1} - 48 < 0$ (4) $32(\frac{1}{4})^x - 18(\frac{1}{2})^x + 1 \le 0$

1. 問題の内容

次の方程式・不等式を解く問題です。

(1) $\sqrt[3]{9^x} = 3 \sqrt[4]{9^x}$

(2) $9^{x+1} + 80 \cdot 3^{x-1} - 1 = 0$

(3) $2^{2x} - 2^{x+1} - 48 < 0$

(4) $32(\frac{1}{4})^x - 18(\frac{1}{2})^x + 1 \le 0$

2. 解き方の手順

(1) $\sqrt[3]{9^x} = 3 \sqrt[4]{9^x}$

両辺を指数で表すと、

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

$9^{\frac{x}{3}} = 3 \cdot 9^{\frac{x}{4}}$

$3^{\frac{2x}{3}} = 3 \cdot 3^{\frac{2x}{4}}$

$3^{\frac{2x}{3}} = 3^{1 + \frac{x}{2}}$

$\frac{2x}{3} = 1 + \frac{x}{2}$

$\frac{2x}{3} - \frac{x}{2} = 1$

$\frac{4x - 3x}{6} = 1$

$\frac{x}{6} = 1$

$x = 6$

(2) $9^{x+1} + 80 \cdot 3^{x-1} - 1 = 0$

$9 \cdot 9^x + 80 \cdot \frac{3^x}{3} - 1 = 0$

$9 \cdot (3^2)^x + \frac{80}{3} \cdot 3^x - 1 = 0$

$9 \cdot (3^x)^2 + \frac{80}{3} \cdot 3^x - 1 = 0$

$3^x = t$ とおくと、

$9t^2 + \frac{80}{3} t - 1 = 0$

両辺に3をかけると、

$27t^2 + 80t - 3 = 0$

$(27t - 1)(t + 3) = 0$

$t = \frac{1}{27}, -3$

$3^x = \frac{1}{27}, -3$

$3^x > 0$ より $3^x = \frac{1}{27} = 3^{-3}$

$x = -3$

(3) $2^{2x} - 2^{x+1} - 48 < 0$

$(2^x)^2 - 2 \cdot 2^x - 48 < 0$

$2^x = t$ とおくと、

$t^2 - 2t - 48 < 0$

$(t - 8)(t + 6) < 0$

$-6 < t < 8$

$-6 < 2^x < 8$

$2^x > 0$ なので、 $0 < 2^x < 8$

$2^x < 2^3$

$x < 3$

(4) $32(\frac{1}{4})^x - 18(\frac{1}{2})^x + 1 \le 0$

$32((\frac{1}{2})^2)^x - 18(\frac{1}{2})^x + 1 \le 0$

$32((\frac{1}{2})^x)^2 - 18(\frac{1}{2})^x + 1 \le 0$

$(\frac{1}{2})^x = t$ とおくと、

$32t^2 - 18t + 1 \le 0$

$(4t - 1)(8t - 1) \le 0$

$\frac{1}{8} \le t \le \frac{1}{4}$

$\frac{1}{8} \le (\frac{1}{2})^x \le \frac{1}{4}$

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

$3 \ge x \ge 2$

$2 \le x \le 3$

3. 最終的な答え

(1) $x = 6$

(2) $x = -3$

(3) $x < 3$

(4) $2 \le x \le 3$