26 pkt za trzy zadania, jeśli do 26.12.16r. Proszę o rozwiązanie do poniższego (nie samą odpowiedź). Zad. 7 Oblicz: a) log_2 80 + log_2 0,1 b) log_27 - log_2 56 Zad. 8 podstaw pod podane wyrażenie jako jeden logarytm: a)log_5 3 +log_5x b) log 8a

7. a) [latex]log_2 80 + log_2 0,1= log_2 (80 cdot 0,1) = log_2 8 = log_2 2^3 = \\ = 3 cdot log_2 2 = 3 cdot 1 = 3 \\ ===== \ Wzory: \ log_a b + log_a c = log_a (b cdot c) \ log_a (b^n) = n cdot log_a b \ log_a a = 1[/latex] b) [latex]log_27 - log_2 56 = log_2 (frac{7}{56})= log_2 (frac{1}{8})= log_2 (frac{1}{2^3})=log_2 (2^{-3})= \\ = -3 cdot log_2 2 = - 3 cdot 1 = -3 \\ Wzory: \ log_a b - log_a c = log_a ( frac{b}{c} ) \ frac{1}{a^n}=a^{-n} \ log_a (b^n) = n cdot log_a b \ log_a a = 1[/latex] 8. a) [latex]log_5 3 +log_5 x = log_5 (3 cdot x) = log_5 (3x)[/latex] b) [latex]log 8a -log 3a - log a = log frac{8a}{3a} - log a= log frac{8}{3} - log a=\\ = log frac{8}{3a}[/latex] c) [latex]3 + log_4 5 = 3 cdot log_4 4 + log_4 5 =log_4 4^3 + log_4 5 = \\ =log_4 64 + log_4 5 =log_4 (64 cdot 5) = log_4 320[/latex] 9. a) [latex]4^x cdot 5^x=20 \\ (4 cdot 5)^x = 20 \\ 20^x = 20^1 \\ x = 1[/latex] b) [latex]3^{x+2} -3^x=24 \\ 3^x cdot 3^2 - 3^x = 24 \\ 3^x cdot 9 - 3^x = 24 \\ 3^x cdot (9 - 1) = 24 \\ 3^x cdot 8 = 24 /:8 \\ 3^x = 3 \\ 3^x = 3^1 \\ x = 1[/latex] c) [latex]12^x=3^x \\ (3 cdot 4)^x=3^x \\ 3^x cdot 4^x=3^x /:3^x \\ 4^x = 1 \\ 4^x = 4^0 \\ x = 0[/latex]