Cześć!
Logarytm:
[tex]log_a(b)=c \iff a^c=b[/tex], gdy [tex]a>0 \ \wedge \ a\not=1 \ \wedge \ b>0[/tex]
Zanim obliczymy wartość ułamka, wyliczmy wartości pojedynczych logarytmów:
[tex]log_5(25) = x \iff 5^x=25 \iff 5^x=5^2 \Longrightarrow x=2\\\\log_{\frac{1}{3}}(27) = y \iff (\frac{1}{3})^y = 27 \iff 3^{-y}=27 \iff 3^{-y}=3^3 \Longrightarrow y=-3[/tex]
Ponadto:
[tex]log_{0,4}(\frac{25}{4}) = log_{\frac{4}{10}}(\frac{100}{16}) = z \iff (\frac{4}{10})^z = \frac{100}{16} \iff (\frac{4}{10})^z =(\frac{16}{100})^{-1} \iff\\\\\iff (\frac{4}{10})^z = (\frac{4}{10})^{-2} \Longrightarrow z=-2\\\\log_{0,4}(\frac{8}{125}) =log_{\frac{4}{10}}(\frac{8}{125}) = a \iff (\frac{4}{10})^a=\frac{8}{125} \iff (\frac{2}{5})^a = \frac{8}{125} \iff\\\\\iff (\frac{2}{5})^a = (\frac{2}{5})^3 \Longrightarrow a=3[/tex]
Zatem:
[tex]\frac{log_{5}(25)\cdot log_{\frac{1}{3}}(27)}{log_{0,4}(\frac{25}{4})\cdot log_{0,4}(\frac{8}{125})} = \frac{x \cdot y}{z \cdot a} =\frac{2 \cdot (-3)}{(-2) \cdot 3} = \frac{-6}{-6}=1[/tex]
Pozdrawiam!
