Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 764

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\[1)\log_{6}x = 3\]

\[\log_{6}x = \log_{6}6^{3}\]

\[x = 6^{3} = 216\]

\[Ответ:\ \ x = 216.\]

\[2)\log_{5}x = 4\]

\[\log_{5}x = \log_{5}5^{4}\]

\[x = 5^{4}\]

\[x = 625\]

\[Ответ:\ \ x = 625.\]

\[3)\log_{2}(5 - x) = 3\]

\[\log_{2}(5 - x) = \log_{2}2^{3}\]

\[5 - x = 2^{3}\]

\[5 - x = 8\]

\[- x = 3\]

\[x = - 3\]

\[Ответ:\ \ x = - 3.\]

\[4)\log_{3}(x + 2) = 3\]

\[\log_{3}(x + 2) = \log_{3}3^{3}\]

\[x + 2 = 3^{3}\]

\[x + 2 = 27\ \]

\[x = 25\]

\[Ответ:\ \ x = 25.\]

\[5)\log_{\frac{1}{6}}(0,5 + x) = - 1\]

\[\log_{\frac{1}{6}}(0,5 + x) = \log_{\frac{1}{6}}\left( \frac{1}{6} \right)^{- 1}\]

\[0,5 + x = \left( \frac{1}{6} \right)^{- 1}\]

\[0,5 + x = 6\ \]

\[x = 5,5\]

\[Ответ:\ \ x = 5,5.\]

\[6)\ \log_{0,2}(3 - x) = - 2\]

\[\log_{0,2}(3 - x) = \log_{0,2}\left( \frac{1}{5} \right)^{- 2}\]

\[3 - x = 25\]

\[x = - 22\]

\[Ответ:\ \ x = - 22.\ \]