与えられた3つの対数を含む式をそれぞれ簡単にします。

1. 問題の内容

与えられた3つの対数を含む式をそれぞれ簡単にします。

2. 解き方の手順

(1) $\log_3{\sqrt{32}} + \log_9{54} - \log_{\sqrt{3}}{6}$

まず、$\sqrt{32} = \sqrt{2^5} = 2^{5/2}$, $\log_{\sqrt{3}}{6} = \log_{3^{1/2}}{6}=2\log_3{6}$に注意します。

また、$\log_9{54} = \frac{\log_3{54}}{\log_3{9}} = \frac{\log_3{(2\cdot 3^3)}}{\log_3{3^2}} = \frac{\log_3{2} + 3\log_3{3}}{2} = \frac{\log_3{2}+3}{2}$

よって、

$\log_3{\sqrt{32}} + \log_9{54} - \log_{\sqrt{3}}{6} = \log_3{2^{5/2}} + \frac{\log_3{2}+3}{2} - 2\log_3{6}$

$= \frac{5}{2}\log_3{2} + \frac{1}{2}\log_3{2} + \frac{3}{2} - 2(\log_3{2}+\log_3{3}) = \frac{6}{2}\log_3{2} + \frac{3}{2} - 2\log_3{2} - 2 = 3\log_3{2} - 2\log_3{2} + \frac{3}{2} - 2 = \log_3{2} - \frac{1}{2}$

$= \log_3{2} - \log_3{\sqrt{3}} = \log_3{\frac{2}{\sqrt{3}}} = \log_3{\frac{2\sqrt{3}}{3}}$

(2) $(\log_4{9} - \log_{16}{3})(\log_9{16} - \log_3{8})$

$\log_4{9} = \frac{\log_2{9}}{\log_2{4}} = \frac{\log_2{3^2}}{2} = \frac{2\log_2{3}}{2} = \log_2{3}$

$\log_{16}{3} = \frac{\log_2{3}}{\log_2{16}} = \frac{\log_2{3}}{4}$

$\log_9{16} = \frac{\log_3{16}}{\log_3{9}} = \frac{\log_3{2^4}}{2} = \frac{4\log_3{2}}{2} = 2\log_3{2}$

$\log_3{8} = \log_3{2^3} = 3\log_3{2}$

$(\log_4{9} - \log_{16}{3})(\log_9{16} - \log_3{8}) = (\log_2{3} - \frac{1}{4}\log_2{3})(2\log_3{2} - 3\log_3{2}) = \frac{3}{4}\log_2{3} (-\log_3{2})$

$= -\frac{3}{4}\log_2{3} \cdot \log_3{2} = -\frac{3}{4}\log_2{3}\cdot \frac{1}{\log_2{3}} = -\frac{3}{4}$

(3) $16^{\log_2{3}}$

$16^{\log_2{3}} = (2^4)^{\log_2{3}} = 2^{4\log_2{3}} = 2^{\log_2{3^4}} = 3^4 = 81$

3. 最終的な答え

(1) $\log_3{\frac{2\sqrt{3}}{3}}$

(2) $-\frac{3}{4}$

(3) $81$