Bài tập 2.49 trang 125 SBT Toán 12

Bài tập 2.49 trang 125 SBT Toán 12

a)

\(\begin{array}{l}
{\log _2}({2^x} + 1).{\log _2}({2^{x + 1}} + 2) = 2\\
 \Leftrightarrow {\log _2}({2^x} + 1).[1 + {\log _2}({2^x} + 1)] = 2\\
 \Leftrightarrow \log _2^2({2^x} + 1) + \log ({2^x} + 1) - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\log _2}({2^x} + 1) = 1\\
{\log _2}({2^x} + 1) =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{2^x} + 1 = 2\\
{2^x} + 1 = \frac{1}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{2^x} = 1\\
{2^x} =  - \frac{3}{4}\,\,\left( l \right)
\end{array} \right. \Leftrightarrow x = 0
\end{array}\)

b) ĐKXĐ: x > 0

\({\log \left( {{x^{\log 9}}} \right) = \log 9.\log x = \log \left( {{9^{\log x}}} \right) \Rightarrow {x^{\log 9}} = {9^{\log x}}}\)

Do đó, ta có:

\(\begin{array}{l}
{x^{\log 9}} + {9^{\log x}} = 6\\
 \Leftrightarrow {x^{\log 9}} = 3\\
 \Leftrightarrow \log 9.\log x = \log 3\\
 \Leftrightarrow \log x = \frac{{\log 3}}{{\log 9}}\\
 \Leftrightarrow \log x = \frac{1}{2}\\
 \Leftrightarrow x = \sqrt {10} 
\end{array}\)

c) ĐKXĐ: x > 0

\(\begin{array}{l}
{x^{3{{\log }^3}x}} - \frac{2}{3}\log x = 100\sqrt[3]{{10}}\\
 \Leftrightarrow \left( {3{{\log }^3}x - \frac{2}{3}\log x} \right)\log x = 2 + \frac{1}{3}\\
 \Leftrightarrow 3{\log ^4}x - \frac{2}{3}{\log ^2}x = \frac{7}{3}\\
 \Leftrightarrow 9{\log ^4}x - 2{\log ^2}x - 7 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\log ^2}x = 1\\
{\log ^2}x =  - 1\,\,\left( l \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\log x = 1\\
\log x =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x = \frac{1}{{10}}
\end{array} \right.
\end{array}\)

 d) ĐKXĐ: x > - 2

\(\begin{array}{l}
1 + 2{\log _{x + 2}}5 = {\log _5}\left( {x + 2} \right)\\
 \Leftrightarrow 1 + 2.\frac{1}{{{{\log }_5}\left( {x + 2} \right)}} = {\log _5}\left( {x + 2} \right)\\
 \Leftrightarrow \log _5^2\left( {x + 2} \right) - {\log _5}\left( {x + 2} \right) - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\log _5}\left( {x + 2} \right) =  - 1\\
{\log _5}\left( {x + 2} \right) = 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = \frac{1}{5}\\
x + 2 = 25
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \frac{9}{5}\\
x = 23
\end{array} \right.
\end{array}\)

-- Mod Toán 12