[latex]log_{frac{1}{16}}2=log_{frac{1}{16}} sqrt[4]{16}=log_{frac{1}{16}} 16^{frac{1}{4}}=log_{frac{1}{16}} left(frac{1}{16} ight)^{-frac{1}{4}}=-frac{1}{4} \ \ log_{32}2sqrt{2}=log_{32}(sqrt[5]{32} cdot sqrt{sqrt[5]{32}})=log_{32} (32^{frac{1}{5}} cdot 32^{frac{1}{10}})=log_{32} 32^{0.3}=0.3 \ \ log 2 - log 0,2 + log 25= log (frac{2}{0.2})+log 25 = log 10+log 25=1+log 25[/latex] log 25 nie da sie obliczyc w pamieci, tak trzeba zostawic Zad 2 a) [latex]log_{frac{1}{3}}x=-3 qquad x extgreater 0 \ \ x=left(frac{1}{3} ight)^{-3}=3^3=27[/latex] b) [latex]log_{x+3} 27=3 qquad x extgreater -3 qquad x eq -2 \ \ (x+3)^3=27 qquad /sqrt[3]{} \ \ x+3=3 \ \ x=0[/latex] c) [latex]log_4 x=2log_4 6 - log_4 frac{9}{4} \ \ log_4 x= log_4 6^2 - log_4 frac{9}{4} \ \ log_4 x=log_4 36 - log_4 frac{9}{4} \ \ log_4 x = log_4 (36 cdot frac{4}{9}) \ \ log_4 x = log_4 16 \ \ x=16[/latex]
Zadanie 1. w zalaczniku. 2. log₁/₃ x=-3 x=(1/3)⁻³ x=3³ x=27 b) log_{x+3}27=3 (x+3)³=27 (x+3)³=3³ x+3=3 x=0 c) log₄x=2log₄6-log₄(9/4) log₄x=log₄6²-log₄(9/4) log₄x=log₄(36:9/4) log₄x=log₄16 x=16
