数学 – 急速・緩慢に変化する関係 - 指数関数・対数関数 – (いくつかの例題及び問題の補充)

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数学読本〈2〉簡単な関数/平面図形と式/指数関数・対数関数/三角関数 (松坂和夫(著)、岩波書店)の第7章(急速・緩慢に変化する関係 - 指数関数・対数関数)、7.3(対数の性質)、いくつかの例題及び問題の補充、問32.を取り組んでみる。

問32.

$\begin{array}{l} t = 3^{x} \\ t^{2} + t - 12 = 0 \\ (t + 4) (t - 3) = 0 \\ t = 3 \\ 3^{x} = 3 \\ x = 1 \end{array}$
$\begin{array}{l} t = 2^{x} \\ 2 t + t^{2} - 80 = 0 \\ (t + 10) (t - 8) = 0 \\ t = 8 \\ 2^{x} = 8 \\ x = 3 \end{array}$
$\begin{array}{l} t = 2^{x} \\ t^{2} - 12 t + 32 = 0 \\ (t - 8) (t - 4) = 0 \\ t = 4, 8 \\ 2^{x} = 4, 8 \\ x = 2, 3 \end{array}$
$\begin{array}{l} t = \log_{10} x \\ t^{2} - 2 t = 0 \\ t (t - 2) = 0 \\ t = 0, 2 \\ \log_{10} x = 0, 2 \\ x = 1, 100 \end{array}$
$\begin{array}{l} t = \log_{2} x \\ t = \frac{1}{t} \\ t^{2} = 1 \\ t = \pm 1 \\ \log_{2} x = \pm 1 \\ x = \frac{1}{2}, 2 \end{array}$
$\begin{array}{l} t = \log_{3} x \\ t^{3} = 4 t \\ t (t^{2} - 4) = 0 \\ t = 0, \pm 2 \\ \log_{3} x = 0, \pm 2 \\ x = 1, 9, \frac{1}{9} \end{array}$

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