Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 251

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\[1)\ 2^{x} + 2^{x - 3} = 18\ \]

\[2^{x} \bullet \left( 1 + 2^{- 3} \right) = 18\ \]

\[2^{x} \bullet \left( \frac{8}{8} + \frac{1}{8} \right) = 18\ \]

\[2^{x} \bullet \frac{9}{8} = 18\ \]

\[2^{x} = 16\ \]

\[2^{x} = 2^{4}\ \]

\[x = 4\ \]

\[Ответ:\ \ x = 4.\]

\[2)\ 3^{x} + 4 \bullet 3^{x + 1} = 13\ \]

\[3^{x} \bullet \left( 1 + 4 \bullet 3^{1} \right) = 13\ \]

\[3^{x} \bullet (1 + 12) = 13\ \]

\[3^{x} \bullet 13 = 13\ \]

\[3^{x} = 1\ \]

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

\[x = 0\ \]

\[Ответ:\ \ x = 0.\]

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

\[3^{x} \bullet \left( 2 \bullet 3^{1} - 6 \bullet 3^{- 1} - 1 \right) = 9\ \]

\[3^{x} \bullet \left( 6 - \frac{6}{3} - 1 \right) = 9\ \]

\[3^{x} \bullet (5 - 2) = 9\ \]

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

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

\[x = 1\ \]

\[Ответ:\ \ x = 1.\]

\[5^{x} \bullet \left( 5^{1} + 3 \bullet 5^{- 1} - 6 \right) = - 10\ \]

\[5^{x} \bullet \left( 5 + \frac{3}{5} - 6 \right) = - 10\ \]

\[5^{x} \bullet \left( \frac{25}{5} + \frac{3}{5} - \frac{30}{5} \right) = - 10\ \]

\[5^{x} \bullet \left( - \frac{2}{5} \right) = - 10\ \]

\[5^{x} = 25\ \]

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

\[x = 2\ \]

\[Ответ:\ \ x = 2.\]