oblicz korzystając z odpowiednich twierdzeń zadanie z załącznikiem pomocy??

a) [latex]log_{0,4}frac{5}{3}+log_{0,4}3-log_{0,4}2=log_{0,4}left(frac{5}{3} cdot 3 ight) -log_{0,4}2=[/latex] [latex]log_{0,4}5 -log_{0,4}2=log_{0,4} frac{5}{2} =[/latex] [latex]log_{0,4}left( frac{2}{5} ight)^{-1} =-log_{ frac{2}{5} }frac{2}{5} =-1[/latex] b) [latex]log_{sqrt2}6+2log_{sqrt2}2+log_{sqrt2}3=log_{sqrt2}6+log_{sqrt2}2^2+log_{sqrt2}3=[/latex] [latex]log_{sqrt2}6+log_{sqrt2}4+log_{sqrt2}3=log_{sqrt2}(6 cdot 4 cdot 3)=[/latex] [latex]log_{sqrt2}72= frac{log72}{logsqrt2}=frac{log72}{log2^{ frac{1}{2} }}=[/latex] [latex]frac{ frac{1}{2} log72}{log2}=frac{2log72}{log2}=[/latex] [latex]frac{2log(2^3 cdot 3^2)}{log2}=frac{2(log(2^3+log3^2))}{log2}=[/latex] [latex]frac{2log2^3+2log3^2}{log2}=frac{2log2^3}{log2}+frac{2log3^2}{log2}=[/latex] [latex]frac{6log2}{log2}+frac{4log3}{log2}=6+4log_23[/latex]