Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 691

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\[1)\ 7^{x - 2} = 3^{2 - x};\]

\[7^{x - 2} = \left( \frac{1}{3} \right)^{- (2 - x)};\]

\[7^{x - 2} = \left( \frac{1}{3} \right)^{x - 2};\]

\[7^{x - 2}\ :\left( \frac{1}{3} \right)^{x - 2} = 1;\]

\[(7 \bullet 3)^{x - 2} = (7 \bullet 3)^{0};\]

\[x - 2 = 0;\]

\[x = 2;\]

\[Ответ:\ \ x = 2.\]

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

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

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

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

\[(2 \bullet 3)^{x - 3} = (2 \bullet 3)^{0};\]

\[x - 3 = 0;\ \]

\[x = 3;\]

\[Ответ:\ \ x = 3.\]

\[3)\ 3^{\frac{x + 2}{4}} = 5^{x + 2};\]

\[\sqrt[4]{3^{x + 2}} = 5^{x + 2};\]

\[\frac{\sqrt[4]{3^{x + 2}}}{5^{x + 2}} = 1;\]

\[\left( \frac{\sqrt[4]{3}}{5} \right)^{x + 2} = \left( \frac{\sqrt[4]{3}}{5} \right)^{0};\]

\[x + 2 = 0;\]

\[x = - 2;\]

\[Ответ:\ \ x = - 2.\]

\[4)\ 4^{\frac{x - 3}{2}} = 3^{2(x - 3)};\]

\[\left( \sqrt{4} \right)^{x - 3} = \left( 3^{2} \right)^{x - 3};\]

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

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

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

\[x - 3 = 0;\ \]

\[x = 3;\]

\[Ответ:\ \ x = 3.\]