ГДЗ по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 11

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\[\textbf{а)}\ 2^{x - 1} = 3^{x}\]

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

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

\[x - 1 = x \cdot \log_{2}3\]

\[x \cdot \log_{2}3 = x - 1\]

\[x \cdot \log_{2}3 - x = - 1\]

\[x \cdot \left( \log_{2}3 - 1 \right) = - 1\]

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

\[\textbf{б)}\ 2^{x} = 3^{x + 1}\]

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

\[2^{x} = 2^{(x + 1) \cdot \log_{2}3}\]

\[x = (x + 1) \cdot \log_{2}3\]

\[x \cdot \log_{2}3 - x = - \log_{2}3\]

\[x \cdot \left( \log_{2}3 - 1 \right) = - \log_{2}3\]

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

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

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

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

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

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

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

\[x - 2 = (x - 3) \cdot \log_{2}3\]

\[x - x \cdot \log_{2}3 = 2 - 3 \cdot \log_{2}3\]

\[x\left( 1 - \log_{2}3 \right) = 2 - 3 \cdot \log_{2}3\]

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

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

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

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

\[x - 3 = (x - 2) \cdot \log_{2}3\]

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

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

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