logarytmy mam do rozwiązania . zdiencia dodaem w zalaczniku.... Question from @muz125wp

Roma

Będziemy korzystać z własności logarytmów:

$log_a \ 1 = 0 \\\ log_a \ a = 1 \\\ log_a \ b^r = r \cdot \log_a \ b \\\ log_a \ (b \cdot c) = log_a \ b + log_a \ c \\\ log_a \ ( \frac{b}{c}) = log_a \ b - log_a \ c$

oraz z własności potęg i pierwiastków:

$(a^r)^s = a^{r \cdot s} \\\ a^{\frac{m}{n}} = \sqrt[n]{a^m} \\\ (\frac{a}{b})^{-n} = (\frac{b}{a})^n$

Zad. 1

$log_3 \ 1 = 0 \\\\ log_3 \ 3 = 1 \\\\ log_3 \ 9 = log_3 \ 3^2 = 2 \cdot log_3 \ 3 = 2 \cdot 1 = 2 \\\\ log_3 \ \sqrt{3} = log_3 \ 3^{\frac{1}{2}} = \frac{1}{2} \cdot log_3 \ 3 = \frac{1}{2} \cdot 1 = \frac{1}{2} \\\\ log_3 \ \frac{1}{3} = log_3 \ 3^{-1} = -1 \cdot log_3 \ 3 = -1 \cdot 1 = -1$

$log_4 \ 1 = 0 \\\\ log_4 \ 4 = 1 \\\\ log_4 \ 16 = log_4 \ 4^2 = 2 \cdot log_4 \ 4 = 2 \cdot 1 = 2 \\\\ log_4 \ 8 = log_4 \ (4 \cdot 2) = log_4 \ 4 + log_4 \ 2 = 1 + log_4 \ \sqrt{4} = \\\ = 1 + log_4 \ 4^{\frac{1}{2}} = 1 + \frac{1}{2} \cdot log_4 \ 4 = 1 + \frac{1}{2} \cdot 1 = 1 + \frac{1}{2} = 1\frac{1}{2} \\\\ log_4 \ 0,5 = log_4 \ \frac{1}{2} =log_4 \ 1 - log_4 \ 2 = 0 - log_4 \ \sqrt{4} = - log_4 \ 4^{\frac{1}{2}} = \\\ = -\frac{1}{2} \cdot log_4 \ 4 = - \frac{1}{2} \cdot 1 = -\frac{1}{2}$

$log_5 \ 1 = 0 \\\\ log_5 \ 5= 1 \\\\ log_5 \ 0,2 = log_5 \ \frac{2}{10} = log_5 \ \frac{1}{5} = log_5 \ 5^{-1} = - 1 \cdot log_5 \ 5 =\\\ = - 1 \cdot 1 = - 1 \\\\ log_5 \ 0,04 = log_5 \ \frac{4}{100} = log_5 \ \frac{1}{25} = log_5 \ \frac{1}{5^2} =log_5 \ 5^{-2} =\\\ = - 2 \cdot log_5 \ 5 = - 2 \cdot 1 = - 2 \\\\ log_5 \ \sqrt{5} = log_5 \ 5^{\frac{1}{2}} =\frac{1}{2} \cdot log_5 \ 5 =\frac{1}{2} \cdot 1 =\frac{1}{2}$

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W treści zadania był do obliczenia $log_5 \ 4$ , ale patrząc na inne przykłady to raczej chodziło o $log_5 \ 5$ i taki logarytm obliczyłam.

Jednak jeśli należało obliczyć $log_5 \ 4$ , to takie logarytmy obliczamy korzystając z tablic logarytmicznych oraz ze wzoru na zmianę podstawy logarytmu:

$log_a \ b = \frac{log_c \ b}{log_c \ a}$

Zatem:

$log_5 \ 4 = \frac{log \ 4}{log \ 5} \approx \frac{0,6}{0,7} \approx 0,86$

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$log_{0,5} \ 1 = 0 \\\\ log_{0,5} \ 0,5 = 1 \\\\ log_{0,5} \ 2 = log_{0,5} \ (\frac{1}{2})^{-1} = - 1 \cdot log_{0,5} \ 0,5 = - 1 \cdot 1 = - 1 \\\\ log_{0,5} \ 8 = log_{0,5} \ 2^3 = log_{0,5} \ (\frac{1}{2})^{-3} = - 3 \cdot log_{0,5} \ 0,5 = - 3 \cdot 1 = - 3$

$log_{0,5} \ \sqrt{32} = log_{0,5} \ 32^{\frac{1}{2}} = log_{0,5} \ (2^5)^{\frac{1}{2}} =log_{0,5} \ 2^{5 \cdot \frac{1}{2}} =log_{0,5} \ 2^{2,5}\\\ =log_{0,5} \ [(\frac{1}{2})^{-1}]^{2,5} =log_{0,5} \ (0,5)^{- 1 \cdot 2,5} = log_{0,5} \ (0,5)^{-2,5} =\\\ = -2,5 \cdot log_{0,5} \ 0,5 = -2,5 \cdot 1 = - 2,5$

$log_{0,1} \ 1 = 0 \\\\ log_{0,1} \ 0,1 = 1 \\\\ log_{0,1} \ 10 = log_{0,1} \ (\frac{1}{10})^{-1} = - 1 \cdot log_{0,1} \ 0,1 = - 1 \cdot 1 = - 1 \\\\ log_{0,1} \ \sqrt {10} = log_{0,1} \ 10^{\frac{1}{2}} = log_{0,1} \ [(\frac{1}{10})^{-1}]^{\frac{1}{2}} = log_{0,1} (0,1)^{-1 \cdot \frac{1}{2}} =\\\ = log_{0,1} (0,1)^{-\frac{1}{2}} =-\frac{1}{2} \cdot log_{0,1} \ 0,1= -\frac{1}{2} \cdot 1 = -\frac{1}{2}$

$log_{0,1} \ 100 = log_{0,1} \ 10^2 = log_{0,1} \ [(\frac{1}{10})^{-1}]^2 = log_{0,1} \ (0,1)^{- 1 \cdot 2} = \\\ = log_{0,1} \ (0,1)^{- 2} = - 2 \cdot log_{0,1} \ 0,1 = - 2 \cdot 1 = - 2$

$log_{\sqrt{2}} \ 1 = 0 \\\\ log_{\sqrt{2}} \ \sqrt{2} = 1 \\\\ log_{\sqrt{2}} \ 2 = log_{\sqrt{2}} \ (\sqrt{2})^2 = 2 \cdot log_{\sqrt{2}} \ \sqrt{2} = 2 \cdot 1 = 2 \\\\ log_{\sqrt{2}} \ 0,25 =log_{\sqrt{2}} \ \frac{25}{100} =log_{\sqrt{2}} \ \frac{1}{4} =log_{\sqrt{2}} \ \frac{1}{2^2} =log_{\sqrt{2}} \ 2^{-2}} =\\\ =log_{\sqrt{2}} \ [(\sqrt{2})^2]^{-2}} =log_{\sqrt{2}} \ (\sqrt{2})^{2 \cdot(-2)}} = log_{\sqrt{2}} \ (\sqrt{2})^{-4}} = \\\ = -4 \cdot log_{\sqrt{2}} \ \sqrt{2} = - 4 \cdot 1 = - 4$

$log_{\sqrt{2}} \ 32 =log_{\sqrt{2}} \ 2^5 = log_{\sqrt{2}} \ [(\sqrt{2})^2]^5= log_{\sqrt{2}} \ (\sqrt{2})^{2 \cdot 5} =\\\ =log_{\sqrt{2}} \ (\sqrt{2})^{10} =10 \cdot log_{\sqrt{2}} \ \sqrt{2} = 10 \cdot 1 = 10$

Zad. 2

$log_a \ x = c \ => \ a^c = x$

$log_2 \ x = 0 \\\ 2^0 = x \\\ x = 1$

$log_2 \ x = 1 \\\ 2^1 = x \\\ x = 2$

$log_2 \ x = 5 \\\ 2^5 = x \\\ x = 32$

$log_2 \ x = 1,5 \\\ 2^{1,5} = x \\\ 2^{\frac{3}{2}} = x \\\ x = \sqrt{2^3} \\\ x = \sqrt{8} \\\ x = \sqrt{4 \cdot 2} \\\ x = 2 \sqrt{2}$

$log_3 \ x = 0 \\\ 3^0 = x \\\ x = 1$

$log_3 \ x = 1 \\\ 3^1 = x \\\ x = 3$

$log_3 \ x = 3 \\\ 3^3 = x \\\ x = 27$

$log_3 \ x = \frac{1}{3} \\\ 3^{\frac{1}{3}} = x \\\ x = \sqrt[3]{3}$

$log_4 \ x = 0 \\\ 4^0 = x \\\ x = 1$

$log_4 \ x = 1 \\\ 4^1 = x \\\ x = 4$

$log_4 \ x = 4 \\\ 4^4 = x \\\ x = 256$

$log_4 \ x = 0,5 \\\ 4^{0,5} = x \\\ x = 4^{\frac{1}{2}} \\\ x = \sqrt{4} \\\ x = 2$

$log_5 \ x = 0 \\\ 5^0 = x \\\ x = 1$

$log_5 \ x = 1 \\\ 5^1 = x \\\ x = 5$

$log_5 \ x = -2 \\\ 5^{-2} = x \\\ x = \frac{1}{5^2} \\\ x = \frac{1}{25}$

$log_5 \ x = - 0,5 \\\ 5^{-0,5} = x \\\ x = 5^{-\frac{1}{2}} \\\ x = \frac{1}{5^{\frac{1}{2}}} \\\ x = \frac{1}{\sqrt{5}} \\\ x =\frac{\sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} \\\ x =\frac{\sqrt{5}}{5}$

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