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1 \({\log _5}x = 3\)
2 \({\log _2}\left( {2x + 1} \right) = 3\)
3 \({\log _2}\left( {x + 1} \right) + {\log _2}\left( {x + 4} \right) = 2\)
4 \({\log _3}243 = 2x + 1\)
5 \({\log _3}\left( {x - 1} \right) = 4\)
6 \(\log \left( {2x} \right) - \log \left( {x - 3} \right) = 1\)
7 \(2{\log _4}\left( {x - 1} \right) = 3\)
1
\({\log _5}x = 3 \Rightarrow x = {5^3} \Rightarrow x = 125\)
2
\({\log _2}\left( {2x + 1} \right) = 3 \Rightarrow 2x + 1 = {2^3} = 8 \Rightarrow 2x = 7 \Rightarrow x = \frac{7}{2}\)
3
\(\begin{array}{l}{\log _2}\left( {x + 1} \right) + {\log _2}\left( {x + 4} \right) = 2\\\\ \Rightarrow {\log _2}\left( {x + 1} \right)\left( {x + 4} \right) = 2\\\\ \Rightarrow {\log _2}\left( {{x^2} + 5x + 4} \right) = 2\\\\ \Rightarrow {x^2} + 5x + 4 = {2^2} = 4\\\\ \Rightarrow {x^2} + 5x = 0 \Rightarrow x\left( {x + 5} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = 0\\{x_2} = - \;5 \otimes \end{array} \right. \Rightarrow x = 0\end{array}\)
4
\(\begin{array}{l}{\log _3}243 = 2x + 1 \Rightarrow {\log _3}{3^5} = 5{\log _3}3\\\\ = 2x + 1 \Rightarrow 5 \times 1 = 2x + 1 \Rightarrow 2x = 4 \Rightarrow x = 2\end{array}\)
5
\({\log _3}\left( {x - 1} \right) = 4 \Rightarrow x - 1 = {3^4} = 81 \Rightarrow x = 82\)
6
\(\begin{array}{l}\log \left( {2x} \right) - \log \left( {x - 3} \right) = 1 \Rightarrow \log \frac{{2x}}{{x - 3}} = 1\\\\ \Rightarrow \frac{{2x}}{{x - 3}} = 10 \Rightarrow 2x = 10x - 30 \Rightarrow x = \frac{{15}}{4}\end{array}\)
7
\(\begin{array}{l}2{\log _4}\left( {x - 1} \right) = 3 \Rightarrow {\log _4}\left( {x - 1} \right) = \frac{3}{2}\\\\ \Rightarrow x - 1 = {4^{\frac{3}{2}}} = 8 \Rightarrow x = 9\end{array}\)
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