Zad 1. A [latex]log_3 sqrt{3} = x[/latex] [latex]3^x = sqrt{3}[/latex] [latex]3^x = 3^{frac{1}{2}}[/latex] [latex]x = frac{1}{2}[/latex] B [latex]log_{sqrt{3}} 3= x[/latex] [latex](sqrt{3})^x = 3[/latex] [latex](3^{frac{1}{2}})^x = 3^1[/latex] [latex]3^{frac{1}{2}x} = 3^1[/latex] [latex]frac{1}{2}x = 1 / cdot 2[/latex] [latex]x = 2[/latex] C [latex]log_4 8 = x[/latex] [latex]4^x = 8[/latex] [latex](2^2)^x = 2^3[/latex] [latex]2^{2x} = 2^3[/latex] [latex]2x = 3 /:2[/latex] [latex]x = frac{3}{2}[/latex] D [latex]log_8 4 = x[/latex] [latex]8^x = 4[/latex] [latex](2^3)^x = 2^2[/latex] [latex]2^{3x} = 2^2[/latex] [latex]3x = 2 /:3[/latex] [latex]x = frac{2}{3}[/latex] Odp. D Zad. 3 [latex](0,6)^{-3} i (frac{5}{3})^2[/latex] [latex](0,6)^{-3} = (frac{6}{10})^{-3} = (frac{3}{5})^{-3} = (frac{5}{3})^3 [/latex] Zatem porównujemy: [latex](frac{5}{3})^3 i (frac{5}{3})^2[/latex] Jeżeli dwie potęgi mają jednakowe podstawy dodatnie większe od 1, to większa jest ta potęga, która ma większy wykładnik. Porównywane potęgi mają jednakowe podstawy i ⁵/₃ = 1⅔ > 1, zatem: [latex](frac{5}{3})^3 > (frac{5}{3})^2[/latex], czyli [latex](0,6)^{-3} > (frac{5}{3})^2[/latex] Zad. 5 a) [latex]log_6 x = 2log_6 4 - log_6 18[/latex] [latex]log_6 x = log_6 4^2 - log_6 18[/latex] [latex]log_6 x = log_6 16 - log_6 18[/latex] [latex]log_6 x = log_6 frac{16}{18}[/latex] [latex]x = frac{16}{18}[/latex] [latex]x = frac{8}{9}[/latex] Odp. x = ⁸/₉ b) [latex]log_{frac{1}{2}} x = - 4[/latex] [latex]x = (frac{1}{2})^{-4}[/latex] [latex]x = 2^4[/latex] [latex]x = 16[/latex] Odp. x = 16 Zad. 6 a) [latex]log_2 frac{1}{8} + log_{frac{1}{16}} 2 = - 3 + (-frac{1}{4}) = -3frac{1}{4}[/latex] ------------------------------- [latex]log_2 frac{1}{8} = x[/latex] [latex]2^x = frac{1}{8}[/latex] [latex]2^x = (frac{1}{2})^3[/latex] [latex]2^x = 2^{-3}[/latex] [latex]x = - 3[/latex] [latex]log_{frac{1}{16}} 2 = x[/latex] [latex](frac{1}{16})^x = 2[/latex] [latex][(frac{1}{2})^4]^x = 2^1[/latex] [latex](2^{-4})^x = 2^1[/latex] [latex]2^{-4x} = 2^1[/latex] [latex]-4x = 1 /:(-4)[/latex] [latex]x = - frac{1}{4}[/latex] b) [latex]36^{log_6 3} = (6^2)^{log_6 3} = 6^{2log_6 3} = 6^{log_6 3^2} = 6^{log_6 9} = 9[/latex] c) [latex]log frac{1}{2} + log 2 = log (frac{1}{2} cdot 2) = log 1 = 0[/latex] ------------------------------- [latex]log 1 = x[/latex] [latex]10^x = 1[/latex] [latex]10^x = 10^0[/latex] [latex]x = 0[/latex]
