OBLICZ log przy podstawie 5 z 5 Log 7z 1 Log 1/3 z 81 tu jest w ułAmku Log 2 z 1/64 Log 1/4 z 1/2 2log 1/5 z 125 2*log 1/3 z 9 mnożenie Log 3 z 1/27 Log 8 z 32+log 8 z2 Log 2 z 4 + log log 2z 8 Log 25+ log 40 Log 5z50-log 5 z 2 Log 2z 24-

[latex]Wzory:\log_{a}b = c to a^{c} = b\log_{a}b + log_{a}c = log_{a}(bcdot c)\log_{a}b - log_{a}c = log_{a}(frac{b}{c})\ncdot log_{a}b = log_{a}(b^{n})[/latex] [latex]log_{5}5 = log_{5}5^{1} = 1\\log_{7}1 = log_{7}7^{0} = 0\\log_{frac{1}{3}}81 = log_{frac{1}{3}}(frac{1}{3})^{-4} = -4\\log_{2}frac{1}{64} = log_{2}2^{-6} = -6\\log_{frac{1}{4}}frac{1}{2}=log_{frac{1}{4}}(frac{1}{4})^{frac{1}{2}}=frac{1}{2}\\2log_{frac{1}{5}}125=log_{frac{1}{5}}125^{2} = log_{frac{1}{5}}((frac{1}{5})^{-3})^{2} = log_{frac{1}{5}}(frac{1}{5})^{-6} = -6[/latex] [latex]2log_{frac{1}{3}}9 = log_{frac{1}{3}}9^{2} = logfrac{1}{3}}((frac{1}{3})^{-2})^{2} = log_{frac{1}{3}}(frac{1}{3})^{-4}} = -4\\log_{3}frac{1}{27} = log_{3}3^{-3}} = -3\\log_{8}32 + log_{8}2 = log_{8}(32cdot2) = log_{8}64 = log_{8}8^{2} = 2\\log_{2}4 + log_{2}8 = log_{2}(4cdot8) = log_{2}32 = log_{2}2^{5} = 5\\log25+log40 = log(25cdot40) = log1000 = log10^{3} = 3\\log_{5}50 - log_{5}2 = log_{5}frac{50}{2} = log_{5}25 = log_{5}5^{2} = 2[/latex] [latex]log_{2}24 - log_{2}3 = log_{2}frac{24}{3} = log_{2}8 = log_{2}2^{3} = 3\\log_{3}36 - log_{3}4 = log_{3}frac{36}{4} = log_{3}9 = log_{3}3^{2} = 2[/latex]