与えられた10個の数式（累乗、指数、対数を含む）の値を求める問題です。

1. 問題の内容

与えられた10個の数式（累乗、指数、対数を含む）の値を求める問題です。

2. 解き方の手順

(1) $\sqrt{8} \div \sqrt[3]{16} \times \sqrt[6]{32}$

$\sqrt{8} = 8^{\frac{1}{2}} = (2^3)^{\frac{1}{2}} = 2^{\frac{3}{2}}$

$\sqrt[3]{16} = 16^{\frac{1}{3}} = (2^4)^{\frac{1}{3}} = 2^{\frac{4}{3}}$

$\sqrt[6]{32} = 32^{\frac{1}{6}} = (2^5)^{\frac{1}{6}} = 2^{\frac{5}{6}}$

よって、

$2^{\frac{3}{2}} \div 2^{\frac{4}{3}} \times 2^{\frac{5}{6}} = 2^{\frac{3}{2} - \frac{4}{3} + \frac{5}{6}} = 2^{\frac{9 - 8 + 5}{6}} = 2^{\frac{6}{6}} = 2^1 = 2$

(2) $2^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times 6^{-\frac{1}{2}}$

$2^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times 6^{-\frac{1}{2}} = 2^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times (2 \times 3)^{-\frac{1}{2}} = 2^{\frac{3}{2}} \times 3^{\frac{5}{2}} \times 2^{-\frac{1}{2}} \times 3^{-\frac{1}{2}} = 2^{\frac{3}{2} - \frac{1}{2}} \times 3^{\frac{5}{2} - \frac{1}{2}} = 2^1 \times 3^2 = 2 \times 9 = 18$

(3) $(9^{\frac{1}{2}})^{\frac{3}{2}} \times 81^{-\frac{1}{4}}$

$(9^{\frac{1}{2}})^{\frac{3}{2}} = 9^{\frac{1}{2} \times \frac{3}{2}} = 9^{\frac{3}{4}} = (3^2)^{\frac{3}{4}} = 3^{\frac{3}{2}}$

$81^{-\frac{1}{4}} = (3^4)^{-\frac{1}{4}} = 3^{-1} = \frac{1}{3}$

よって、

$3^{\frac{3}{2}} \times \frac{1}{3} = 3^{\frac{3}{2} - 1} = 3^{\frac{1}{2}} = \sqrt{3}$

(4) $64^2 \div 4^{-2} \times 8^{-\frac{16}{3}}$

$64^2 = (2^6)^2 = 2^{12}$

$4^{-2} = (2^2)^{-2} = 2^{-4}$

$8^{-\frac{16}{3}} = (2^3)^{-\frac{16}{3}} = 2^{-16}$

よって、

$2^{12} \div 2^{-4} \times 2^{-16} = 2^{12 - (-4) - 16} = 2^{12 + 4 - 16} = 2^0 = 1$

(5) $\log_2{\sqrt{8}}$

$\sqrt{8} = 8^{\frac{1}{2}} = (2^3)^{\frac{1}{2}} = 2^{\frac{3}{2}}$

$\log_2{\sqrt{8}} = \log_2{2^{\frac{3}{2}}} = \frac{3}{2}$

(6) $8^{\log_2{5}}$

$8^{\log_2{5}} = (2^3)^{\log_2{5}} = 2^{3 \log_2{5}} = 2^{\log_2{5^3}} = 5^3 = 125$

(7) $\log_5{75} - \log_5{15}$

$\log_5{75} - \log_5{15} = \log_5{\frac{75}{15}} = \log_5{5} = 1$

(8) $\log_3{\sqrt{5}} - \frac{1}{2} \log_3{10} + \log_3{\sqrt{18}}$

$\log_3{\sqrt{5}} - \frac{1}{2} \log_3{10} + \log_3{\sqrt{18}} = \log_3{\sqrt{5}} - \log_3{\sqrt{10}} + \log_3{\sqrt{18}} = \log_3{\frac{\sqrt{5} \times \sqrt{18}}{\sqrt{10}}} = \log_3{\frac{\sqrt{5 \times 18}}{\sqrt{10}}} = \log_3{\sqrt{\frac{5 \times 18}{10}}} = \log_3{\sqrt{9}} = \log_3{3} = 1$

(9) $(\log_{16}{3} - \log_4{9})(\log_3{16} + \log_9{4})$

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

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

$\log_3{16} = \frac{\log_2{16}}{\log_2{3}} = \frac{4}{\log_2{3}}$

$\log_9{4} = \frac{\log_3{4}}{\log_3{9}} = \frac{\log_3{2^2}}{2} = \frac{2\log_3{2}}{2} = \log_3{2} = \frac{\log_2{2}}{\log_2{3}} = \frac{1}{\log_2{3}}$

$(\frac{\log_2{3}}{4} - \log_2{3})(\frac{4}{\log_2{3}} + \frac{1}{\log_2{3}}) = (\frac{1}{4} - 1)\log_2{3} \times (\frac{4+1}{\log_2{3}}) = -\frac{3}{4} \log_2{3} \times \frac{5}{\log_2{3}} = -\frac{3}{4} \times 5 = -\frac{15}{4}$

(10) $(3^{\frac{2}{3}} - 2^{\frac{2}{3}})(3^{\frac{2}{3}} + 3^{\frac{1}{3}} \times 2^{\frac{2}{3}} + 2^{\frac{4}{3}})(3+2)$

これは $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ の形に似ています。

ここで、$a = 3^{\frac{1}{3}}$、$b = 2^{\frac{1}{3}}$とすると、

$a^2 = (3^{\frac{1}{3}})^2 = 3^{\frac{2}{3}}$, $b^2 = (2^{\frac{1}{3}})^2 = 2^{\frac{2}{3}}$, $ab = 3^{\frac{1}{3}}2^{\frac{1}{3}}$, $a^3 = (3^{\frac{1}{3}})^3 = 3$, $b^3 = (2^{\frac{1}{3}})^3 = 2$

$(3^{\frac{2}{3}} - 2^{\frac{2}{3}})(3^{\frac{2}{3}} + 3^{\frac{1}{3}} \times 2^{\frac{2}{3}} + 2^{\frac{4}{3}}) = (a-b)((a^2) + ab+ (b)^2)$

$a^3 - b^3=(a-b)(a^2+ab+b^2)$, $a=(3^{1/3}), a^2 = (3^{2/3})$,$b = 2^{\frac{2}{3}}$

$(3^{\frac{1}{3}}) - = a^3 = 5$

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

3. 最終的な答え

(1) 2

(2) 18

(3) $\sqrt{3}$

(4) 1

(5) $\frac{3}{2}$

(6) 125

(7) 1

(8) 1

(9) $-\frac{15}{4}$

(10) 5