Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 695

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

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

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

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

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

\[x + 3 = 0;\]

\[x = - 3;\]

\[Ответ:\ \ x = - 3.\]

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

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

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

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

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

\[x - 2 = 0;\]

\[x = 2;\]

\[Ответ:\ \ x = 2.\]

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

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

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

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

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

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

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

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

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

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

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

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

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

\[4x - 4 = 9;\]

\[4x = 13;\]

\[x = 3,25;\]

\[Выражение\ имеет\ смысл\ при:\]

\[x - 1 \geq 0;\]

\[x \geq 1.\]

\[Ответ:\ \ x = 3,25.\ \]