Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 498

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

\[7^{x\sqrt{3}} = 7^{\frac{1}{2}};\]

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

\[x = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{2 \bullet 3} = \frac{\sqrt{3}}{6};\]

\[Ответ:\ \ x = \frac{\sqrt{3}}{6}.\]

\[2)\ 25^{x\sqrt{2}} = 5\sqrt{5};\]

\[\left( 5^{2} \right)^{x\sqrt{2}} = 5^{1} \bullet 5^{\frac{1}{2}};\]

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

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

\[x = \frac{3}{2 \bullet 2\sqrt{2}} = \frac{3\sqrt{2}}{2 \bullet 2 \bullet 2} = \frac{3\sqrt{2}}{8};\]

\[Ответ:\ \ x = \frac{3\sqrt{2}}{8}.\]

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

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

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

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

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

\[Ответ:\ \ x = 1.\]