معادله های لگاریتمی زیر را حل کنید:
الف \({\log _5}\left( {2x - 1} \right) = {\log _5}x\)
ب \({\log _3}\left( {x - 1} \right) + {\log _3}\left( {\frac{x}{2} + 1} \right) = 2\)
پ \(\log x + \log \left( {x + 3} \right) = 1\)
الف
\(\begin{array}{l}\left\{ \begin{array}{l}2x - 1 > 0 \Rightarrow x > \frac{1}{2}\quad \left( i \right)\\x > 0\quad \left( {ii} \right)\end{array} \right.\quad \mathop \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \;x > \frac{1}{2}\quad \left( {iii} \right)\\{\log _5}\left( {2x - 1} \right) = {\log _5}x \Rightarrow 2x - 1 = x\;\mathop \Rightarrow \limits^{\left( {iii} \right)} \;x = 1\end{array}\)
ب
\(\begin{array}{l}\left\{ \begin{array}{l}x - 1 > 0 \Rightarrow x > 1\quad \left( i \right)\\\frac{x}{2} + 1 > 0 \Rightarrow x > - 2\quad \left( {ii} \right)\end{array} \right.\quad \mathop \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \;x > 1\quad \left( {iii} \right)\\\\{\log _3}\left( {x - 1} \right) + {\log _3}\left( {\frac{x}{2} + 1} \right) = 2\\\\ \Rightarrow {\log _3}\left\{ {\left( {x - 1} \right)\left( {\frac{x}{2} + 1} \right)} \right\} = 2\\\\ \Rightarrow \frac{1}{2}{x^2} + \frac{1}{2}x - 1 = 9 \Rightarrow {x^2} + x - 20 = 0\\\\ \Rightarrow \left( {x - 4} \right)\left( {x + 5} \right) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}x = - 5\\x = 4\end{array} \right.\;\mathop \Rightarrow \limits^{\left( {iii} \right)} \;x = 4\end{array}\)
پ
\(\begin{array}{l}\left\{ \begin{array}{l}x > 0\quad \left( i \right)\\x + 3 > 0 \Rightarrow x > - 3\quad \left( {ii} \right)\end{array} \right.\quad \mathop \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \;x > 0\quad \left( {iii} \right)\\\\\log x + \log \left( {x + 3} \right) = 1 \Rightarrow \log \left\{ {x\left( {x + 3} \right)} \right\} = 1\\\\ \Rightarrow {x^2} + 3x = 10\\\\ \Rightarrow {x^2} + 3x - 10 = 0 \Rightarrow \left( {x + 5} \right)\left( {x - 2} \right) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}x = - 5\\x = 2\end{array} \right.\;\mathop \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \;x = 2\end{array}\)