Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 327

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\[1)\log_{3}(5x - 1) = 2\]

\[\log_{3}(5x - 1) = \log_{3}3^{2}\]

\[5x - 1 = 3^{2}\]

\[5x - 1 = 9\]

\[5x = 10\ \]

\[x = 2\]

\[Ответ:\ \ x = 2.\]

\[2)\log_{5}(3x + 1) = 2\]

\[\log_{5}(3x + 1) = \log_{5}5^{2}\]

\[3x + 1 = 5^{2}\]

\[3x + 1 = 25\]

\[3x = 24\ \]

\[x = 8\]

\[Ответ:\ \ x = 8.\]

\[3)\log_{4}(2x - 3) = 1\]

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

\[2x - 3 = 4\]

\[2x = 7\ \]

\[x = 3,5\]

\[Ответ:\ \ x = 3,5.\]

\[4)\log_{7}(x + 3) = 2\]

\[\log_{7}(x + 3) = \log_{7}7^{2}\]

\[x + 3 = 7^{2}\]

\[x + 3 = 49\ \]

\[x = 46\]

\[Ответ:\ \ x = 46.\]

\[5)\lg(3x - 1) = 0\]

\[\lg(3x - 1) = \lg 1\]

\[3x - 1 = 1\]

\[3x = 2\]

\[x = \frac{2}{3}\]

\[Ответ:\ \ x = \frac{2}{3}.\]

\[6)\lg(2 - 5x) = 1\]

\[\lg(2 - 5x) = \lg 10\]

\[2 - 5x = 10\]

\[- 5x = 8\]

\[x = - 1,6\]

\[Ответ:\ \ x = - 1,6.\]