Question

Poproszę o rozwiązanie zadania 84 i 86.

Answer 1

Zadanie 84

[tex]3log_{9} 3 - log_{9}6 + 2log_{9} 2 + log_{9} 4,5 =\\log_{9}3^{3} - log_{9}6 + log_{9} 2^{2} + log_{9} 4,5 = \\log_{9} 27 - log_{9}6 + log_{9} 4 + log_{9} 4,5 = \\log_{9}(\frac{27*4*4,5}{6}) = log_{9}(\frac{162}{2}) = log_{9}81 = log_{9}9^{2} = 2log_{9}9 = 2[/tex]

Zadanie 86

[tex](3log_{\frac{4}{3} }4 - 2log_{\frac{4}{3} }6)^{2} =\\(log_{\frac{4}{3} }4^{3} - log_{\frac{4}{3} }6^{2} )^{2}=\\(log_{\frac{4}{3} }64 - log_{\frac{4}{3} }36 )^{2}=\\(log_{\frac{4}{3} }\frac{64}{36}) ^{2} = (log_{\frac{4}{3} }\frac{16}{9}) ^{2} = (log_{\frac{4}{3} }(\frac{4}{3})^{2}) ^{2} = 2^{2} = 4[/tex]

Answer 2

Szczegółowe wyjaśnienie:

Logarytmy - najważniejsze wzory

Jeżeli a > 0, a ≠ 1, b > 0 oraz c > 0, to zachodzą następujące wzory:

[tex]log_{a}b + log_{a}c = log_{a}(b\cdot c)\\\\log_{a}b - log_{a}c = log_{a}(\frac{b}{c})\\\\n\cdot log_{a}b = log_{a}(b^{n})[/tex]

84.

[tex]3log_{9}3 - log_{9}6 + 2log_{9}2 + log_{9}4,5 = \\\\=log_{9}3^{3}-log_{9}6 +log_{9}2^{2}+log_{9}4,5=\\\\=log_{9}27-log_{9}6 + log_{9}4+log_{9}4,5=\\\\=log_{9}\frac{27}{6}+log_{9}(4\cdot4,5)=\\\\=log_{9}\frac{9}{2}+log_{9}18=\\\\=log_{9}(\frac{9}{2}\cdot18)\\\\=log_{9}81=\\\\=log_{9}9^{2}=\\\\=2log_{9}9 = \boxed{2}[/tex]

86.

[tex](3log_{\frac{4}{3}}4-2log_{\frac{4}{3}}6)^{2}=\\\\=(log_{\frac{4}{3}}4^{3} - log_{\frac{4}{3}}6^{2})^{2}=\\\\=(log_{\frac{4}{3}}64-log_{\frac{4}{3}}36)^{2}=\\\\=(log_{\frac{4}{3}}\frac{64}{36})^{2}}=\\\\=(log_{\frac{4}{3}}\frac{16}{9})^{2}=\\\\=(log_{\frac{4}{3}}(\frac{4}{3})^{2}})^{2}=\\\\=(2log_{\frac{4}{3}}\frac{4}{3})^{2}=\\\\=2^{2} =\boxed{ 4}[/tex]