a) \(\displaystyle {\log _2}({2^x} + 1).{\log _2}\left[ {2({2^x} + 1)} \right] = 2\)\(\displaystyle \Leftrightarrow {\log _2}({2^x} + 1).\left[ {1 + {{\log }_2}({2^x} + 1)} \right] = 2\)
Đặt \(\displaystyle t = {\log _2}({2^x} + 1)\), ta có phương trình \(\displaystyle t\left( {1 + t} \right) = 2\; \Leftrightarrow {t^2} + t - 2 = 0\)\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = 1\\t = - 2\end{array} \right.\)
\(\displaystyle \Rightarrow \left[ \begin{array}{l}{\log _2}({2^x} + 1) = 1\\{\log _2}({2^x} + 1) = - 2\end{array} \right.\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}{2^x} + 1 = 2\\{2^x} + 1 = \frac{1}{4}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{2^x} = 1\\{2^x} = - \frac{3}{4}(l)\end{array} \right.\)\(\displaystyle \Leftrightarrow x = 0\)
b) ĐK: \(\displaystyle x > 0\).
Ta có: \(\displaystyle \log ({x^{\log 9}}) = \log 9.\log x\) và \(\displaystyle \log ({9^{\log x}}) = \log x.\log 9\)
Nên \(\displaystyle \log ({x^{\log 9}}) = \log ({9^{\log x}})\) suy ra \(\displaystyle {x^{\log 9}} = {9^{\log x}}\)
Đặt \(\displaystyle t = {x^{\log 9}}\), ta được phương trình \(\displaystyle 2t = 6 \Leftrightarrow t = 3\) \(\displaystyle \Leftrightarrow {x^{\log 9}} = 3\)
\(\displaystyle \Leftrightarrow \log ({x^{\log 9}}) = \log 3\)\(\displaystyle \Leftrightarrow \log 9.\log x = \log 3\)
\(\displaystyle \Leftrightarrow \log x = \frac{{\log 3}}{{\log 9}}\)\(\displaystyle \Leftrightarrow \log x = \frac{1}{2}\)
\(\displaystyle \Leftrightarrow x = \sqrt {10} \) (thỏa mãn điều kiện \(\displaystyle x > 0\))
c) ĐK: \(\displaystyle x > 0\).
Lấy logarit thập phân hai vế của phương trình đã cho, ta được: \(\displaystyle (3{\log ^3}x - \frac{2}{3}\log x).\log x = \frac{7}{3}\)
Đặt \(\displaystyle t = \log x\), ta được phương trình \(\displaystyle 3{t^4} - \frac{2}{3}{t^2} - \frac{7}{3} = 0\)
\(\displaystyle \Leftrightarrow 9{t^4} - 2{t^2} - 7 = 0\)\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}{t^2} = 1\\{t^2} = - \frac{7}{9}(l)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}t = 1\\t = - 1\end{array} \right.\) \(\displaystyle \Rightarrow \left[ \begin{array}{l}\log x = 1\\\log x = - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 10\\x = \frac{1}{{10}}\end{array} \right.\).
d) Đặt \(\displaystyle t = {\log _5}(x + 2)\) với điều kiện \(\displaystyle x + 2 > 0,x + 2 \ne 1\), ta có: \(\displaystyle 1 + \frac{2}{t} = t\)\(\displaystyle \Leftrightarrow {t^2} - t - 2 = 0,t \ne 0\)
\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = - 1\\t = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{\log _5}(x + 2) = - 1\\{\log _5}(x + 2) = 2\end{array} \right.\)\(\displaystyle \Leftrightarrow \left[ \begin{array}{l}x + 2 = \frac{1}{5}\\x + 2 = 25\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{9}{5}\\x = 23\end{array} \right.\).
