Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 219

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

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

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

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

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

\[x - 2 = 0\ \]

\[x = 2\]

\[Ответ:\ \ x = 2.\]

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

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

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

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

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

\[x - 3 = 0\]

\[x = 3\]

\[Ответ:\ \ x = 3.\]

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

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

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

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

\[x + 2 = 0\]

\[x = - 2\]

\[Ответ:\ \ x = - 2.\]

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

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

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

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

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

\[x - 3 = 0\ \]

\[x = 3\]

\[Ответ:\ \ x = 3.\]