第10章 10.23

(1)

$\log(x-3) + \log x = 1 \qquad \log x(x-3) = 1 \qquad x(x-3) = 10 \qquad x^2 - 3x - 10 = 0$

$(x-5)(x+2) = 0 \qquad x = -2, 5 \qquad \text{対数の真数は正であることから} \qquad x = 5$

(2)

$\log(x^2 - 6x + 8) = \log(x-2) + 1 \qquad \log\frac{x^2-6x+8}{x-2} = 1 \qquad \log\frac{(x-4)(x-2)}{x-2} = 1$

$\log(x-4) = 1 \qquad x - 4 = 10 \qquad x = 14$

(3)

$\log(3-x) + \log(x+3) = 0 \qquad 3 - x = \frac{1}{x+3} \qquad 9 - x^2 = 1 \qquad x^2 = 8 \qquad x = \pm 2\sqrt{2}$

(4)

$\log_2 x + \log_8 x = 2(\log_2 x)(\log_8 x) \qquad \log_2 x + \frac{\log_2 x}{\log_2 8} = 2(\log_2 x)\left(\frac{\log_2 x}{\log_2 8}\right)$

$\log_2 x + \frac{\log_2 x}{3} = 2(\log_2 x)\left(\frac{\log_2 x}{3}\right) \qquad \frac{4}{3}\log_2 x - \frac{2}{3}\log_2 x \cdot \log_2 x = 0$

$2\log_2 x\,(2 - \log_2 x) = 0 \qquad \log_2 x = 0 \qquad \log_2 x = 2 \qquad x = 1, 4$