Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 858

\[\boxed{\mathbf{858}.}\]

\[\log_{2}\left( 2^{x} + 1 \right)\log_{2}\left( 2^{x + 1} + 2 \right) = 2\]

\[Область\ определения:\]

\[2^{x} + 1 > 0;\ \ \ \ \ \ 2^{x + 1} + 2 > 0\]

\[x \in ( - \infty; + \infty).\]

\[\log_{2}\ \left( 2^{x} + 1 \right)\ \log_{2}\ \left( 2 \cdot 2^{x} + 2 \right) =\]

\[= 2\]

\[\log_{2}\ \left( 2^{x} + 1 \right)\ \log_{2}\ \left( 2 \cdot \left( 2^{x} + 1 \right) \right) = 2\]

\[\log_{2}\ \left( 2^{x} + 1 \right)\left( \log_{2}\ 2 + \ \log_{2}\ \left( 2^{x} + 1 \right) \right) = 2\]

\[\log_{2}\ \left( 2^{x} + 1 \right)\left( 1 + \log_{2}\ \left( 2^{x} + 1 \right) \right) = 2\]

\[\log_{2}\ \left( 2^{x} + 1 \right) = t:\]

\[t \cdot (t + 1) = 2\]

\[t^{2} + t - 2 = 0\]

\[t_{1} + t_{2} = - 1;\ \ \ \ t_{1} \cdot t_{2} = - 2\]

\[t_{1} = - 2;\ \ \ t_{2} = 1.\]

\[\textbf{а)}\ \log_{2}\ \left( 2^{x} + 1 \right) = - 2\]

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

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

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

\[корней\ нет.\]

\[\textbf{б)}\ \log_{2}\ \left( 2^{x} + 1 \right) = 1\]

\[2^{x} + 1 = 2^{1}\]

\[2^{x} = 1\]

\[2^{x} = 2^{0}\]

\[x = 0.\]

\[Ответ:x = 0.\]