a) \(\frac{1}{2}\log \left( {{x^2} + x - 5} \right) = \log 5x + \log \frac{1}{{5x}}.\)
Điều kiện: \(\left\{ \begin{array}{l}{x^2} + x - 5 > 0\\5x > 0\\\frac{1}{{5x}} > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x > \frac{{ - 1 + \sqrt {21} }}{2}\\x < \frac{{ - 1 - \sqrt {21} }}{2}\end{array} \right.\\x > 0\end{array} \right. \Leftrightarrow x > \frac{{ - 1 + \sqrt {21} }}{2} \approx 1,79.\)
\(\begin{array}{l}Pt \Leftrightarrow \frac{1}{2}\log \left( {{x^2} + x - 5} \right) = \log \left( {5x.\frac{1}{{5x}}} \right)\\ \Leftrightarrow \frac{1}{2}\log \left( {{x^2} + x - 5} \right) = \log 1\\\Leftrightarrow \log \left( {{x^2} + x - 5} \right) = 0\\\Leftrightarrow {x^2} + x - 5 = {10^0}=1\\ \Leftrightarrow {x^2} + x - 6 = 0\\\Leftrightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 3 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 3\;\;\left( {ktm} \right)\\x = 2\;\;\left( {tm} \right)\end{array} \right..\end{array}\)
Vậy phương trình có nghiệm \(x=2\).
b) \(\frac{1}{2}\log \left( {{x^2} - 4x - 1} \right) = \log 8x - \log 4x.\)
Điều kiện: \(\left\{ \begin{array}{l}{x^2} - 4x - 1 > 0\\8x > 0\\4x > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x > 2 + \sqrt 5 \\x < 2 - \sqrt 5 \end{array} \right.\\x > 0\end{array} \right. \Leftrightarrow x > 2 + \sqrt 5 .\)
\(\begin{array}{l}Pt \Leftrightarrow \frac{1}{2}\log \left( {{x^2} - 4x - 1} \right) = \log \frac{{8x}}{{4x}}\\ \Leftrightarrow \log \sqrt {{x^2} - 4x - 1} = \log 2\\ \Leftrightarrow \sqrt {{x^2} - 4x - 1} = 2\\ \Leftrightarrow {x^2} - 4x - 1 = 4\\ \Leftrightarrow {x^2} - 4x - 5 = 0\\\Leftrightarrow \left( {x + 1} \right)\left( {x - 5} \right) = 0\\\Leftrightarrow \left[ \begin{array}{l}x + 1 = 0\\x - 5 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 1\;\;\left( {ktm} \right)\\x = 5\;\;\left( {tm} \right)\end{array} \right..\end{array}\)
Vậy phương trình có nghiệm \(x=5.\)
c) \({\log _{\sqrt 2 }}x + 4{\log _4}x + {\log _8}x = 13.\)
Điều kiện: \(x > 0.\)
\(\begin{array}{l}Pt \Leftrightarrow {\log _{{2^{\frac{1}{2}}}}}x + 4{\log _{{2^2}}}x + {\log _{{2^3}}}x = 13\\\Leftrightarrow 2{\log _2}x + 4.\frac{1}{2}{\log _x}x + \frac{1}{3}{\log _2}x = 13\\\Leftrightarrow \frac{{13}}{3}{\log _2}x = 13\\\Leftrightarrow {\log _2}x = 3\\\Leftrightarrow x = {2^3} = 8\;\;\left( {tm} \right).\end{array}\)
Vậy phương trình có nghiệm \(x=8.\)
