与えられた10個の対数方程式を解き、$x$の値を求める問題です。

1. 問題の内容

与えられた10個の対数方程式を解き、$x$の値を求める問題です。

2. 解き方の手順

対数関数の定義 $log_a b = c \Leftrightarrow a^c = b$ を利用して、各方程式を解きます。

(1) $log_2 x = 2$

$x = 2^2 = 4$

(2) $log_5 x = -1$

$x = 5^{-1} = \frac{1}{5}$

(3) $log_3 x = \frac{1}{2}$

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

(4) $log_{10} x = -\frac{1}{2}$

$x = 10^{-\frac{1}{2}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$

(5) $log_{\frac{1}{4}} x = 2$

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

(6) $log_3 (x-1) = 2$

$x - 1 = 3^2 = 9$

$x = 9 + 1 = 10$

(7) $log_2 (3x - 1) = -2$

$3x - 1 = 2^{-2} = \frac{1}{4}$

$3x = \frac{1}{4} + 1 = \frac{5}{4}$

$x = \frac{5}{4} \cdot \frac{1}{3} = \frac{5}{12}$

(8) $log_2 (2x + 5) = \frac{1}{2}$

$2x + 5 = 2^{\frac{1}{2}} = \sqrt{2}$

$2x = \sqrt{2} - 5$

$x = \frac{\sqrt{2} - 5}{2}$

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

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

$-2x = \frac{1}{4} - 3 = \frac{1}{4} - \frac{12}{4} = -\frac{11}{4}$

$x = \frac{11}{4} \cdot \frac{1}{2} = \frac{11}{8}$

(10) $log_{0.2} (2 - 3x) = -1$

$2 - 3x = (0.2)^{-1} = (\frac{1}{5})^{-1} = 5$

$-3x = 5 - 2 = 3$

$x = -1$

3. 最終的な答え

(1) $x = 4$

(2) $x = \frac{1}{5}$

(3) $x = \sqrt{3}$

(4) $x = \frac{\sqrt{10}}{10}$

(5) $x = \frac{1}{16}$

(6) $x = 10$

(7) $x = \frac{5}{12}$

(8) $x = \frac{\sqrt{2} - 5}{2}$

(9) $x = \frac{11}{8}$

(10) $x = -1$