Odpowiedź:
[tex]a)\ \ \dfrac{\sqrt[3]{25}\cdot\frac{1}{125}\cdot\sqrt[4]{5}}{5^3}=\dfrac{\sqrt[3]{5^2}\cdot(\frac{1}{5})^3\cdot5^{\frac{1}{4}}}{5^3}=\dfrac{5^{\frac{2}{3}}\cdot5^{-3}\cdot5^{\frac{1}{4}}}{5^3}=\dfrac{5^{\frac{2}{3}-3+\frac{1}{4}}}{5^3}=\dfrac{5^{\frac{2}{3}-\frac{3}{1}+\frac{1}{4}}}{5^3}=\\\\\\=\dfrac{5^{\frac{8}{12}-\frac{36}{12}+\frac{3}{12}} }{5^3}=\dfrac{5^{-\frac{25}{12}}}{5^3}=5^{-\frac{25}{12}-3}=5^{-\frac{25}{12}-\frac{36}{12}}=5^{-\frac{61}{12}}[/tex]
[tex]b)\ \ \dfrac{\sqrt[4]{128}\cdot32}{\frac{1}{4}\cdot\sqrt[4]{8}}=\dfrac{\sqrt[4]{2^7}\cdot2^5}{(\frac{1}{2})^2\cdot\sqrt[4]{2^3}}=\dfrac{2^{\frac{7}{4}}\cdot2^5}{2^{-2}\cdot2^{\frac{3}{4}}}=\dfrac{2^{\frac{7}{4}+5}}{2^{-2+\frac{3}{4}}}=\dfrac{2^{\frac{7}{4}+\frac{20}{4}}}{2^{-\frac{8}{4}+\frac{3}{4}}}=\dfrac{2^{\frac{27}{4}} }{2^{-\frac{5}{4}}}=2^{\frac{27}{4}-(-\frac{5}{4})}}=\\\\=2^{\frac{27}{4}+\frac{5}{4}}=2^\frac{32}{4}}=2^8[/tex]
[tex]c)\ \ \sqrt{2\sqrt[3]{16}}=\sqrt{2\sqrt[3]{2^4}}=\sqrt{2\cdot2^{\frac{4}{3}}}=\sqrt{2^{1+\frac{4}{3}}}=\sqrt{2^\frac{7}{3}}}=(2^{\frac{7}{3}})^{\frac{1}{2}}=2^{\frac{7}{3}\cdot\frac{1}{2}}=2^{\frac{7}{6}}[/tex]
[tex]Zastosowane\ \ wzory\\\\\sqrt[n]{a^m}=a^{\frac{m}{n}}\\\\a^m\cdot a^n=a^{m+n}\\\\a^m:a^n=a^{m-n}[/tex]
[tex]Oblicz\\\\a)\ \ log_{2\sqrt{2}}32=log_{2\cdot2^{\frac{1}{2}}}32=2_{2^{1+\frac{1}{2}}}32=2_{2^{\frac{3}{2}}}2^5=\frac{5}{\frac{3}{2}}\cdot log_{2}2=5:\frac{3}{2}\cdot1=5\cdot\frac{2}{3}=\\\\=\frac{10}{3}=3\frac{1}{3}[/tex]
[tex]b)\ \ log_{4}128-log_{4}512=log_{2^2}2^7-log_{2^2}2^9=\frac{7}{2}\cdot log_{2}2-\frac{9}{2}\cdot log_{2}2=\frac{7}{2}\cdot1-\frac{9}{2}\cdot1=\\\\=\frac{7}{2}-\frac{9}{2}=-\frac{2}{2}=-1[/tex]
[tex]c)\ \ 2log_{3}\sqrt[4]{9}+log_{3}\frac{1}{81}=2log_{3}\sqrt[4]{3^2}+log_{3}(\frac{1}{3})^4=2log_{3}\sqrt{3}+log_{3}3^{-4}=\\\\=2log_{3}3^{\frac{1}{2}}+log_{3}3^{-4}=\not2\cdot\frac{1}{\not2}+(-4)=1-4=-3[/tex]
[tex]d)\ \ 7^{2+log_{7}14}=7^2\cdot7^{log_{7}14}=49\cdot14=686[/tex]
[tex]Zastosowane\ \ wzory\\\\log_{a^{y}}a^{x}=\frac{x}{y}\\\\log_{a}a^{x}=x\\\\a^{log_{a}b}=b[/tex]
