nalezy wykorzustac def. logarytmu 27^x=9 3^(3x)=3^2 3x=2 x=2/3 ------------------------- 25^x=1/5 5^(2x)=5^(-1) 2x=-1 x=-1/2 ------------------------------------- 69^x=1/69 69^x=69^(-1) x=-1 ---------------------------- 4^x=1/64 4^x=1/[4^3] 4^x=4^(-3) x=-3 pozdr Hans
[latex]log_{27}9=x\27^x=9\(3^3)^x=3^2\3x=2\x=frac{2}{3}\log_{25}frac{1}{5}=x\25^x=frac{1}{5}\5^{2x}=5^{-1}\2x=-1\x=-frac{1}{2}\log_{69}frac{1}{69}=-1\bo\69^{-1}=frac{1}{69}\log_4frac{1}{64}=-3\bo\4^{-3}=frac{1}{64}[/latex]
Oblicz: a)[latex]log 1000[/latex] b)[latex]log100+log _{4} frac{1}{8} [/latex] c)[latex]log _frac{1}{3} 3[/latex] d)[latex]log _{0,5} frac{1}{128} [/latex]...
Oblicz: a) [latex]log 1000[/latex] b) [latex]log100 + log _{4} frac{1}{8} [/latex] c) [latex]log_frac{1}{3} 3_[/latex] d) [latex]log _{0,5} frac{1}{128} [/latex]...
oblicz: a) [latex]log_3 {x} =2[/latex] b) [latex]log_5 {x} =1[/latex] c) [latex]log_1_1 {x} =0[/latex] d) [latex]log_{x} = frac{1}{3} [/latex] e) [latex]log_1 {x}=3[/latex] f) [latex]log_4{x} =3[/latex]...
oblicz a)[latex]log_{2}6+log_{2}12+log_{2}frac{4}{9}[/latex] b)[latex](7^{sqrt{2})sqrt{2}[/latex] c) [latex](frac{16}{81})^{frac{1}{4}}[/latex]...
Wiedząc że:a) [latex]log _{5} 4=a i log _{5}27=b, oblicz log _{5} 108 oraz log _{5} frac{27}{64} [/latex]b) [latex]log _{4} 3=a i log _{4} 5 =b, oblicz log _{4} 0,6 oraz log _{4} 5 frac{2}{5} [/latex]...
