Przyjmij, że log 4 ≈ 0,602 i oblicz: a) log 400 b) log 0,4 c) log 0,004 d) log 400 000 e) log(4*10do potęgi 12ej) f)log (4*10do potęgi -9)   Prosże o pomoc w rozwiązaniu, próbowałem z pkt. a jednak mój wynik różnił się z tymi na końcu strony. Pozdrawiam.

[latex]log 4 approx 0,602 [/latex]   a) [latex]log 400 = log (4 cdot 100) = log 4 + log 100 = 0,602 + 2 = 2,602[/latex] ___________________ [latex]log 100 = x [/latex] [latex]10^x = 100[/latex] [latex]10^x = 10^2[/latex] [latex]x = 2[/latex] [latex]log 100 = 2[/latex]     b) [latex]log 0,4 = log frac{4}{10} = log 4 - log 10 = 0,602 - 1 = - 0,398[/latex]   __________________ [latex]log 10 = x[/latex] [latex]10^x = 10[/latex] [latex]10^x = 10^1[/latex] [latex]x = 1[/latex] [latex]log 10 = 1[/latex]     c) [latex]log 0,004 = log frac{4}{1000} = log 4 - log 1000 = 0,602 - 3 = - 2,398[/latex] _________________ [latex]log 1000 = x[/latex] [latex]10^x = 1000[/latex] [latex]10^x = 10^3[/latex] [latex]x = 3[/latex] [latex]log 1000 = 3[/latex]     d) [latex]log 400000= log (4 cdot 100000) = log 4 + log 100000 = 0,602 + 5 = 5,602[/latex] ____________________ [latex]log 100000 = x[/latex] [latex]10^x = 100000[/latex] [latex]10^x = 10^5[/latex] [latex]x = 5[/latex] [latex]log 100000 = 5[/latex]     e) [latex]log(4 cdot 10^{12}) = log 4 + log 10^{12} = 0,602 + 12 = 12,602[/latex] _____________________ [latex]log 10^{12} = x[/latex] [latex]10^x = 10^{12}[/latex] [latex]x =12[/latex] [latex]log 10^{12} = 12[/latex]     f) [latex]log (4 cdot 10^{-9}) = log 4 + log 10^{-9} = 0,602 + (- 9) = 0,602 - 9 = - 8,398[/latex] ______________________ [latex] log 10^{-9} = x[/latex] [latex]10^x = 10^{-9} [/latex] [latex]x = - 9[/latex] [latex] log 10^{-9} = - 9[/latex]