[tex]1) \ \frac{5^{8}\cdot2^{9}}{10^{7}} = \frac{5^{8}\cdot2^{8}\cdot2}{10^{7}}=2\cdot\frac{(5\cdot2)^{8}}{10^{7}} = 2\cdot\frac{10^{8}}{10^{7}} = 2\cdot10^{8-7} = 2\cdot10^{1} = 2\cdot10 = \boxed{20}[/tex]
[tex]2) \ \frac{6^{-14}+12\cdot(\frac{1}{6})^{15}}{0,5^{14}\cdot(\frac{1}{3})^{15}} = \frac{6^{-14}\cdot2\cdot6\cdot6^{-15}}{(\frac{1}{2})^{14}\cdot(\frac{1}{3})^{14}\cdot\frac{1}{3}} = \frac{6^{-14}+2\cdot6^{-14}}{2^{-14}\cdot3^{-14}\cdot3^{-1}}=\frac{6^{-14}(1+2)}{(2\cdot3)^{-14}\cdot3^{-1}}=\frac{6^{-14}\cdot3}{6^{-14}\cdot3^{-1}} =\\\\= 3^{1-(-1)} = 3^{2}=\boxed{9}[/tex]
[tex]3) \ 36log_{2}\sqrt[4]{2\sqrt[3]{2}}}=36log_{2} (2^{\frac{1}{3}})^{\frac{1}{4}}\cdot2^{\frac{1}{4}} = 36log_{2} 2^{\frac{1}{12}}\cdot2^{\frac{1}{4}} = 36log_{2} 2^{\frac{1}{12}+\frac{1}{4}} = 36log_{2}2^{\frac{1}{12}+\frac{3}{12}} =\\=36log_{2}2^{\frac{4}{12}} = 36log_{2} 2^{\frac{1}{3}} = 36\cdot\frac{1}{3}log_{2} 2 = 12log_{2}2 =\boxed{ 12}[/tex]
[tex]4) \ 2log_{3}6 - log_{3}4 = log_{3}6^{2}-log_{3}4 = log_{3}36-log_{3}4 = log_{3}\frac{36}{4} = log_{3}9 = log_{3}3^{2} =\\\\= 2log_{3}3 =\boxed{2}[/tex]
[tex]5) \ 9^{1-log_{3}6} = 9^{1}:9^{log_{3}6} = 9:(3^{2})^{log_{3}6} =9:3^{2log_{3}6} = 9:3^{log_{3}6^{2}} = 9:6^{2} =\\\\= \frac{9}{36} =\boxed{ \frac{1}{4}}[/tex]
Użyte wzory:
[tex]a^{n}\cdot a^{m} = a^{n+m}\\\\a^{n}:a^{m} = a^{n+m}\\\\(a^{m})^{n} = a^{m\cdot n}\\\\(\frac{a}{b})^{-n} = (\frac{b}{a})^{n}\\\\nlog_{a}b = log_{a}b^{n}\\\\a^{log_{a}b} = b[/tex]
