Rozwiązanie w załączniku.
[latex]log_{0,1} 2+log_{0,1} 2,5+log_{0,1} 20=\ log_{0,1}(2cdot2,5cdot20)=\ log_{0,1}100=\ log_{0,1}(0,1)^{-2}=\ -2[/latex] [latex]dfrac{log_432}{log_48}= dfrac{dfrac{log_232}{log_24}}{dfrac{log_28}{log_24}}= dfrac{log_232}{log_24}cdotdfrac{log_24}{log_2 8}= 5cdotdfrac{1}{3}= dfrac{5}{3}[/latex]
LOGARYTMY!! Bardzo pilne! Oblicz: [latex](log_sqrt{7} [/latex][latex]6 + log_{sqrt{7}}[/latex][latex]frac{1}{42} ) ( log_{frac{1}{2}} 1frac{1}{8} - log_{frac{1}{2}} 9 ) [/latex] ...
1. Oblicz podane logarytmy i uporządkuj je rosnąco: a=[latex]log_{3}[/latex]9 y=[latex]log_{16}frac{1}{4}[/latex] z= [latex]log_{frac{1}{5}}25[/latex] t=[latex]log_{4}2[/latex] ...
Logarytmy; Oblicz: a) [latex]log _{2/3} frac{81}{16} [/latex] b) [latex]log _{1/3} X=-1/2[/latex] c) [latex]log_{2} X=-2/3[/latex]...
OBLICZ (logarytmy) a) [latex] frac{log_3 27 - log_3 1}{3^-^2 * 3^8} = [/latex] b) [latex] frac{log_2 16 + log_2 1}{2^-^3 * ( frac{1}{4})^-^2 } = [/latex]...
logarytmy. POMOCY 1. Liczbę [latex]{3log _2}3-{2log _2}5[/latex] można zapisać w postaci 2. Oblicz [latex]{log _ frac{1}{3} }3 sqrt{3}[/latex] 3. Oblicz [latex]{log _{2 sqrt{2}} } frac{1}{4}[/latex]...
