ГДЗ по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 16

Авторы:
Год:2020-2021-2022
Тип:учебник

Задание 16

\[\boxed{\mathbf{16.}}\]

\[\textbf{а)}\log_{11}\left( \frac{x + 20}{x} \right) = \frac{\log_{2}41}{\log_{2}11}\]

\[\log_{11}\left( \frac{x + 20}{x} \right) = \log_{11}41\]

\[\frac{x + 20}{x} = 41\]

\[41x = x + 20;\ \ \ x \neq 0\]

\[40x = 20\]

\[x = \frac{1}{2}.\]

\[Проверка:\]

\[\log_{11}\left( \left( \frac{1}{2} + 20 \right) \cdot 2 \right) = \log_{11}41\]

\[\log_{11}41 = \log_{11}41\]

\[x = \frac{1}{2} - корень.\]

\[Ответ:x = \frac{1}{2}.\]

\[\textbf{б)}\log_{13}\left( \frac{x + 11}{x} \right) = \frac{\log_{3}23}{\log_{3}13}\]

\[\log_{13}\left( \frac{x + 11}{x} \right) = \log_{13}23\]

\[\frac{x + 11}{x} = 23\]

\[23x = x + 11;\ \ x \neq 0\]

\[22x = 11\]

\[x = \frac{1}{2}.\]

\[Проверка:\]

\[\log_{13}\left( \left( \frac{1}{2} + 11 \right) \cdot 2 \right) = \log_{13}23\]

\[\log_{13}23 = \log_{13}23\]

\[x = \frac{1}{2} - корень.\]

\[Ответ:x = \frac{1}{2}.\]

\[\textbf{в)}\log_{5}\left( \frac{x + 10}{x} \right) = \frac{\log_{11}21}{\log_{11}5}\]

\[\log_{5}\left( \frac{x + 10}{x} \right) = \log_{5}21\]

\[\frac{x + 10}{x} = 21\]

\[21x = x + 10;\ \ x \neq 0\]

\[20x = 10\]

\[x = \frac{1}{2}.\]

\[Проверка:\]

\[\log_{5}\left( \left( \frac{1}{2} + 10 \right) \cdot 2 \right) = \log_{5}21\]

\[\log_{5}21 = \log_{5}21\]

\[x = \frac{1}{2} - корень.\]

\[Ответ:x = \frac{1}{2}.\]

\[\textbf{г)}\log_{7}\left( \frac{x + 15}{x} \right) = \frac{\log_{13}31}{\log_{13}7}\]

\[\log_{7}\left( \frac{x + 15}{x} \right) = \log_{7}31\]

\[\frac{x + 15}{x} = 31\]

\[31x = x + 15;\ \ x \neq 0\]

\[30x = 15\]

\[x = \frac{1}{2}.\]

\[Проверка:\]

\[\log_{7}\left( \left( \frac{1}{2} + 15 \right) \cdot 2 \right) = \log_{7}31\]

\[\log_{7}31 = \log_{7}31\]

\[x = \frac{1}{2} - корень.\]

\[Ответ:x = \frac{1}{2} - корень.\]

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