Question

Rezolvare detaliată, mulțumesc anticipat !

Answer 1

Răspuns:

Explicație pas cu pas:

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Answer 2

$\textbf{Formule:}$

$\log_{a^x}b = \log_{a}b^{\frac{1}{x}}\\ \log_{a}b^x = \log_{a^\frac{1}{x}}b\\a^{\log_{a}b} = b\\\log_{a}b = \dfrac{1}{\log_{b}a}$

$a)\quad \log_{2}\left(\log_{3}\left(\log_{4}64\right)\right) =\log_{2}\left(\log_{3}\left(\log_{4}4^3\right)\right)=\\ \\ = \log_{2}\left(\log_{3}\left(3\log_{4}4\right)\right)=\log_{2}\left(\log_{3}(3\cdot 1)\right) =\\ \\ =\log_{2}(\log_{3}3) = \log_{2}1 = 0$

$b)\quad \log_{\frac{1}{2}}\left[\log_{\sqrt{6}}\left(\log_{3}729\right)\right]=\log_{\frac{1}{2}}\left[\log_{\sqrt{6}}\left(\log_{3}9^3\right)\right]=\\\\ =\log_{\frac{1}{2}}\left[\log_{\sqrt{6}}\left(3\log_{3}9\right)\right] = \log_{\frac{1}{2}}\left[\log_{\sqrt{6}}\left(3\cdot 2\right)\right]=\\ \\ =\log_{\frac{1}{2}}\left(\log_{6}6^2\right) = \log_{\frac{1}{2}}(2\log_{6}6) = \log_{2^{-1}}2=\\ \\ = \log_{2}2^{-1} = -\log_{2}2 = -1$

$c)\quad \log_{\frac{49}{4}}\left(\log_{4}6^{\log_{6}128}\right) =\log_{\frac{49}{4}}\left(\log_{4}128\right) =\\ \\=\log_{(\frac{7}{2})^2}\left(\log_{4}(64\cdot 2 )\right) =\log_{(\frac{7}{2})^2}\left(\log_{4}64+\log_{4}2\right) =\\ \\=\log_{(\frac{7}{2})^2}\left(3+\frac{1}{\log_{2}4} \right)= \log_{(\frac{7}{2})^2}\left(3+\frac{1}{2}\right) = \log_{(\frac{7}{2})^2}\left(\frac{7}{2}\right) =\\ \\ = \log_{\frac{7}{2}}(\frac{7}{2})^{\frac{1}{2}} = \frac{1}{2}$

$d)\quad \log_{\sqrt{2}-1}(\sqrt{3}+1)+\log_{\sqrt{2}+1}(\sqrt{3}+1) = \\ \\ =\log_{(\sqrt{2}+1)^{-1}}(\sqrt{3}+1)+\log_{\sqrt{2}+1}(\sqrt{3}+1) \\ \\ = \log_{\sqrt{2}+1}(\sqrt{3}+1)^{-1}+\log_{\sqrt{2}+1}(\sqrt{3}+1) \\ \\ =\log_{\sqrt{2}+1}\left[(\sqrt{3}+1)^{-1}\cdot (\sqrt{3}+1)\right]\\ \\ = \log_{\sqrt{2}+1}1\\ \\ = 0$