Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 225

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\[1)\ 3^{2x + 6} = 2^{x + 3}\]

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

\[9^{x + 3} = 2^{x + 3}\]

\[\frac{9^{x + 3}}{2^{x + 3}} = 1\]

\[\left( \frac{9}{2} \right)^{x + 3} = \left( \frac{9}{2} \right)^{0}\]

\[x + 3 = 0\]

\[x = - 3.\]

\[Ответ:\ \ x = - 3.\]

\[2)\ 5^{x - 2} = 4^{2x - 4}\]

\[5^{x - 2} = 4^{2(x - 2)}\]

\[5^{x - 2} = 16^{x - 2}\]

\[\frac{5^{x - 2}}{16^{x - 2}} = 1\]

\[\left( \frac{5}{16} \right)^{x - 2} = \left( \frac{5}{16} \right)^{0}\]

\[x - 2 = 0\ \]

\[x = 2.\]

\[Ответ:\ \ x = 2.\]

\[3)\ 2^{x} \bullet 3^{x} = 36^{x^{2}}\]

\[(2 \bullet 3)^{x} = 36^{x^{2}}\]

\[6^{x} = 6^{2x^{2}}\]

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

\[2x^{2} - x = 0\]

\[x(2x - 1) = 0\]

\[x_{1} = 0;\ \text{\ \ }x_{2} = \frac{1}{2} = 0,5.\]

\[Ответ:\ \ x_{1} = 0;\ \ \ x_{2} = 0,5.\]

\[4)\ 9^{- \sqrt{x - 1}} = \frac{1}{27}\]

\[\left( \frac{1}{9} \right)^{\sqrt{x - 1}} = \left( \frac{1}{3} \right)^{3}\]

\[\left( \frac{1}{3} \right)^{2\sqrt{x - 1}} = \left( \frac{1}{3} \right)^{3}\]

\[2\sqrt{x - 1} = 3\]

\[4(x - 1) = 9\]

\[4x - 4 = 9\]

\[4x = 13x = 3,25.\]

\[Выражение\ имеет\ смысл:\]

\[x - 1 \geq 0\ \]

\[x \geq 1.\]

\[Ответ:\ \ x = 3,25.\ \]