Giải bài 2.68 trang 133 - SBT Giải tích lớp 12
Giải các phương trình sau :
\(\begin{align} & a)\,\ln \left( 4x+2 \right)-\ln \left( x-1 \right)=\ln x; \\ & b)\,{{\log }_{2}}\left( 3x+1 \right){{\log }_{3}}x=2{{\log }_{2}}\left( 3x+1 \right); \\ & c)\,{{2}^{{{\log }_{3}}{{x}^{2}}}}{{.5}^{{{\log }_{3}}x}}=400; \\ & d)\,{{\ln }^{3}}x-3{{\ln }^{2}}x-4\ln x+12=0. \\ \end{align} \)
a)
ĐK: \(\left\{ \begin{aligned} & 4x+2>0 \\ & x-1>0 \\ & x>0 \\ \end{aligned} \right.\Leftrightarrow x>1 \)
\(\begin{aligned} & \ln \left( 4x+2 \right)-\ln \left( x-1 \right)=\ln x \\ & \Leftrightarrow \ln \left( 4x+2 \right)=\ln x+\ln \left( x-1 \right) \\ & \Rightarrow 4x+2=x\left( x-1 \right) \\ & \Leftrightarrow {{x}^{2}}-5x-2=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{5+\sqrt{33}}{2} \\ & x=\dfrac{5-\sqrt{33}}{2}\,\,\,\,\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ \end{aligned}\)
b)
ĐK: \(\left\{ \begin{aligned} & 3x+1>0 \\ & x>0 \\ \end{aligned} \right.\Leftrightarrow x>0 \)
\(\begin{aligned} & {{\log }_{2}}\left( 3x+1 \right){{\log }_{3}}x=2{{\log }_{2}}\left( 3x+1 \right) \\ & \Leftrightarrow {{\log }_{2}}\left( 3x+1 \right)\left( {{\log }_{3}}x-2 \right)=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}\left( 3x+1 \right)=0 \\ & {{\log }_{3}}x=2 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & 3x+1=1 \\ & x={{3}^{2}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=0\,\,\,\,\,\,\left( \text{loại} \right) \\ & x=9 \\ \end{aligned} \right. \\ \end{aligned} \)
c)
ĐK: \( x>0 \)
\(\begin{aligned} & {{2}^{{{\log }_{3}}{{x}^{2}}}}{{.5}^{{{\log }_{3}}x}}=400 \\ & \Leftrightarrow {{2}^{2{{\log }_{3}}x}}{{.5}^{{{\log }_{3}}x}}=400 \\ & \Leftrightarrow {{20}^{{{\log }_{3}}x}}={{20}^{2}} \\ & \Leftrightarrow {{\log }_{3}}x=2 \\ & \Leftrightarrow x=9 \,\text{(thỏa mãn ĐK)} \\ \end{aligned}\)
d)
ĐK: \(x> 0\)
\(\begin{aligned} & {{\ln }^{3}}x-3{{\ln }^{2}}x-4\ln x+12=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \ln x=-2 \\ & \ln x=3 \\ & \ln x=2 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x={{e}^{-2}} \\ & x={{e}^{3}} \\ & x={{e}^{2}} \\ \end{aligned} \right. \\ \end{aligned} \)