Zadanie z logarytmów, pilnie potrzebuję pomocy. Zadanie jest w załączniku. Zaznaczone jako "Zadanie domowe"

Zamieniasz liczby logarytmowane na potęgi opodstawie  podstawy logarytmu: a) [latex]log_{3}frac{27}{sqrt[4]{243}}=log_{3}27-log_{3}sqrt[4]{243}=log_{3}3^{3}-log_{3}sqrt[4]{3^{5}} = \ 3-log_{3}3^{frac{5}{4}}=3-frac{5}{4}=1frac{3}{4}[/latex] b) [latex]log_{frac{1}{2}} frac{256sqrt{2}}{sqrt[3]{2}}=log_{frac{1}{2}} 256+log_{frac{1}{2}}sqrt{2}-log_{frac{1}{2}}sqrt[3]{2}=\ =log_{frac{1}{2}}2^{8}+log_{frac{1}{2}}2^{frac{1}{2}}-log_{frac{1}{2}}2^{frac{1}{3}}=log_{frac{1}{2}} (frac{1}{2})^{-8} + log_{frac{1}{2}} (frac{1}{2})^{-frac{1}{2}}-log_{frac{1}{2}} (frac{1}{2})^{-frac{1}{3}}=\ =-8-frac{1}{2}+frac{1}{3}=-8frac{1}{6}[/latex] c) [latex]log_{frac{1}{3}} frac{81sqrt[3]{3}}{sqrt[4]{3}}= log_{frac{1}{3}} 81+log_{frac{1}{3}} sqrt[3]{3}-log_{frac{1}{3}}sqrt[4]{3} = \ = log_{frac{1}{3}} (frac{1}{3})^{-4}+log_{frac{1}{3}} (frac{1}{3})^{-frac{1}{3}}-log_{frac{1}{3}} (frac{1}{3})^{-frac{1}{4}}=-4-frac{1}{3}+frac{1}{4}= \ =-4frac{1}{12}[/latex] d) [latex]log_{5} frac{25sqrt[3]{5}}{sqrt[4]{125}}=log_{5}25 + log_{5}sqrt[3]{5}-log_{5}sqrt[4]{5^{3}}= \ =log_{5}5^{2}+log_{5}5^{frac{1}{3}}-log_{5}5^{frac{3}{4}}=2+frac{1}{3}-frac{3}{4}=1frac{3}{12}+frac{4}{12}=1frac{7}{12}[/latex] e) [latex]log_{frac{1}{4}} frac{2sqrt[5]{64}}{sqrt{8}}=log_{frac{1}{4}}2+log_{frac{1}{4}} sqrt[5]{4^{3}}-log_{frac{1}{4}}sqrt{4^{3}}= \ = log_{frac{1}{4}}4^{frac{1}{2}}+log_{frac{1}{4}}4^{frac{3}{5}}-log_{frac{1}{4}}4^{frac{3}{2}}=log_{frac{1}{4}} (frac{1}{4})^{-frac{1}{2}}+log_{frac{1}{4}} (frac{1}{4})^{-frac{3}{5}}- \ -log_{frac{1}{4}}(frac{1}{4})^{-frac{3}{2}}=-frac{1}{2}-frac{3}{5}+frac{3}{2}=1-frac{3}{5}=frac{2}{5}[/latex] f) [latex]log_{6} frac{sqrt[3]{36}}{216}=log_{6} sqrt[3]{6^{2}}-log_{6}216=log_{6}6^{frac{2}{3}}-log_{6}6^{3}=frac{2}{3}-3=-2frac{1}{3}[/latex] Ze wzorów: [latex]log_{a}(bc)=log_{a}b+log_{a}c \ log_{a}frac{b}{c}=log_{a}b-log_{a}c \ log_{a}a^{b}=b \ a^{-x}=(frac{1}{a})^{x} \ sqrt[n]{a^{m}}=a^{frac{m}{n}} qquad n eq 0 \ [/latex]