2)
Sumę, czy różnicę logarytmów o różnych podstawach możemy obliczyć na dwa sposoby:
I. korzystając ze wzoru na zamianę podstawy:
[tex]log_{a}b = \frac{log_{c}b}{log_{c}a}, \ \ gdzie: \ b \in R+, \ \ a,c \ \in R+ \setminus\{1\}[/tex]
II. korzystając z definicji logarytmu obliczamy poszczególne logarytmy później je sumując lub w przykladzie b) odejmując:
[tex]log_{a}b = c \ \ to \ \ a^{c} = b, \ \ a > 0 \ i \ a \neq 1, \ \ b > 0[/tex]
[tex]I.\\log_{\frac{1}{4}}8 + log_{8}\frac{1}{4} = \frac{log_{2}8}{log_{2}\frac{1}{4}} + \frac{log_{2}\frac{1}{4}}{log_{2}8} = \frac{log_{2}2^{3}}{log_{2}2^{-2}}+\frac{log_{2}2^{-2}}{log_{2}2^{3}} = \frac{3}{-2} + \frac{-2}{3} = -\frac{3}{2}-\frac{2}{3}=\\\\=-\frac{9}{6}-\frac{4}{6} = -\frac{13}{6} = -2\frac{1}{6}[/tex]
[tex]II.[/tex]
[tex]log_{\frac{1}{4}}8 =x\\\\(\frac{1}{4})^{x} = 8\\\\(2^{-2})^{x} = 2^{3}\\\\2^{-2x} = 2^{3}\\\\-2x = 3 \ /:(_2)\\\\x = -\frac{3}{2}\\\\log_{\frac{1}{4}}8 = -\frac{3}{2}[/tex]
[tex]log_{8}\frac{1}{4} = x\\\\8^{x} = \frac{1}{4}\\\\(2^{3})^{x} = 2^{-2}\\\\2^{3x} = 2^{-2}\\\\3x = -2 \ \ /:3\\\\x = -\frac{2}{3}\\\\log_{8}\frac{1}{4} = -\frac{2}{3}[/tex]
[tex]log_{\frac{1}{4}}8 + log_{8}+\frac{1}{4} = -\frac{3}{2}+(-\frac{2}{3}) = -\frac{9}{6}-\frac{4}{6} = -\frac{13}{6} = -2\frac{1}{6}[/tex]
b)
II.
[tex]log_{2}\sqrt[3]{4}-log_{\sqrt[3]{2}}4\\---------\\\\log_{2}\sqrt[3]{4} = x\\\\2^{x} = \sqrt[3]{4}\\\\2^{x} = (2^{2})^{\frac{1}{3}}\\\\2^{x} = 2^{\frac{2}{3}}\\\\log_{2}\sqrt[3]{4} = \frac{2}{3}[/tex]
[tex]log_{\sqrt[3]{2}}4 = x\\\\(\sqrt[3]{2})^{x} = 4\\\\(2^{\frac{1}{3}})^{x} = 2^{2}\\\\2^{\frac{1}{3}x} = 2^{2}\\\\\frac{1}{3}x = 2 \ \ /\cdot3\\\\x = 6\\\\log_{\sqrt[3]{2}}4 = 6[/tex]
[tex]log_{2}\sqrt[3]{4} - log_{\sqrt[3]{2}}4 = \frac{2}{3}-6 = -5\frac{1}{3}[/tex]
