Решите уравнение \(\log_{\sqrt{x + 1}}|5x - 1| = 4\log_{|5x - 1|} \sqrt{x + 1}.\)

Решение

Обозначим за \(t\) выражение \(\log_{\sqrt{x + 1}}|5x - 1|\). Тогда исходное уравнение перепишется в виде:
\[t = \frac{4}{t} \Rightarrow \frac{t^2 - 4}{t} = 0 \Rightarrow t = \pm2.\]
Отсюда:
\begin{gather*}
\left[\begin{array}{l}
\log_{\sqrt{x + 1}}|5x - 1|= 2,\\
\log_{\sqrt{x + 1}}|5x - 1| = -2. \\
\end{array}\right.\Leftrightarrow \ \\
\Leftrightarrow \begin{cases}
\left[\begin{gathered}
x + 1 = |5x \hspace{0.1cm}- 1|,\\
\frac{1}{x \hspace{0.1cm}- 1} = |5x \hspace{0.1cm}- 1|,
\end{gathered}
\right.
\\x + 1 > 0, \\
x + 1 \neq 1.
\end{cases} \Leftrightarrow \ \\
\Leftrightarrow
\begin{cases}
\left[\begin{array}{l}
x + 1 = 5x \hspace{0.1cm}- 1,\\
x + 1 = 1 \hspace{0.1cm}- 5x,\\
(x + 1)(5x \hspace{0.1cm}- 1) = 1,\\
(x + 1)(1 \hspace{0.1cm}- 5x) = 1,\\
\end{array}
\right.
\\x > \hspace{0.1cm}-1, \\
x \neq 0.
\end{cases} \Leftrightarrow \ \\
\Leftrightarrow
\begin{cases}
\left[\begin{array}{l}
x = \frac12,\\
x = 0,\\
5x^2 + 4x \hspace{0.1cm}- 2 = 0,\\
5x^2 + 4x = 0,\\
\end{array}
\right.
\\x > \hspace{0.1cm}-1, \\
x \neq 0.
\end{cases} \Leftrightarrow \ \\
\Leftrightarrow
\begin{cases}
\left[\begin{array}{l}
x = \frac12,\\
x = 0,\\
x = \frac{-2 \pm \sqrt{14}}{5},\\
x = \hspace{0.1cm}-\frac45,\\
\end{array}
\right.
\\x > \hspace{0.1cm}-1, \\
x \neq 0.
\end{cases} \Leftrightarrow \ \\
\Leftrightarrow
\left[\begin{array}{l}
x = \frac12,\\
x = \frac{-2 + \sqrt{14}}{5},\\
x = \hspace{0.1cm}-\frac45.\\
\end{array}
\right.
\end{gather*}

Ответ: \(-\frac{4}{5}\), \(\frac 12\), \(\frac{-2 + \sqrt{14}}{5}\).