Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1348

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\[1)\ 5^{x + 1} + 5^{x} + 5^{x - 1} = 155\]

\[5^{x} \bullet \left( 5^{1} + 5^{0} + 5^{- 1} \right) = 155\]

\[5^{x} \bullet \left( 5 + 1 + \frac{1}{5} \right) = 155\]

\[5^{x} \bullet \frac{25 + 5 + 1}{5} = 155\]

\[5^{x} \bullet \frac{31}{5} = 155\]

\[5^{x} = 25\]

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

\[x = 2.\]

\[Ответ:\ \ x = 2.\]

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

\[3^{2x} \bullet \left( 3^{0} - 2 \bullet 3^{- 1} - 2 \bullet 3^{- 2} \right) = 1\]

\[3^{2x} \bullet \left( 1 - \frac{2}{3} - \frac{2}{9} \right) = 1\]

\[3^{2x} \bullet \frac{9 - 6 - 2}{9} = 1\]

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

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

\[2x = 2\]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]

\[3)\ 7^{x} - 7^{x - 1} = 6\]

\[7^{x} \bullet \left( 7^{0} - 7^{- 1} \right) = 6\]

\[7^{x} \bullet \left( 1 - \frac{1}{7} \right) = 6\]

\[7^{x} \bullet \frac{6}{7} = 6\]

\[7^{x} = 7\]

\[x = 1.\]

\[Ответ:\ \ x = 1.\]

\[4)\ 3^{x + 2} + 3^{x} = 10\]

\[3^{x} \bullet \left( 3^{2} + 3^{0} \right) = 10\]

\[3^{x} \bullet (9 + 1) = 10\]

\[3^{x} \bullet 10 = 10\]

\[3^{x} = 1\]

\[3^{x} = 3^{0}\]

\[x = 0.\]

\[Ответ:\ \ x = 0.\]