Câu 23: \(4^{2x-3}=5\)
=>\(2x-3=log_45=\dfrac{1}{2}\cdot log_25\)
=>\(2x=\dfrac{1}{2}\cdot log_25+3\)
=>\(x=\dfrac{1}{4}\cdot log_25+\dfrac{3}{2}\)
=>Chọn C
Câu 24:
\(7^{2x-1}+2m^2-m=3\)
=>\(7^{2x-1}=-2m^2+m+3\)
Để phương trình này có nghiệm thì \(-2m^2+m+3>0\)
=>\(2m^2-m-3< 0\)
=>\(2m^2-3m+2m-3< 0\)
=>(2m-3)(m+1)<0
=>\(-1< m< \dfrac{3}{2}\)
=>Chọn A
Câu 26:
\(9^x-6^x=2^{2x+1}\)
=>\(9^x-2\cdot4^x-6^x=0\)
=>\(\left(\dfrac{3}{2}\right)^{2x}-\left(\dfrac{3}{2}\right)^x-2=0\)
=>\(\left[\left(\dfrac{3}{2}\right)^x-2\right]\left[\left(\dfrac{3}{2}\right)^x+1\right]=0\)
=>\(\left(\dfrac{3}{2}\right)^x-2=0\)
=>\(x=log_{1,5}2>0\)
=>Chọn B
Câu 27:
\(3^x\cdot2^{x^2}=1\)
=>\(log_3\left(3^x\cdot2^{x^2}\right)=log_31\)
=>\(x+x^2\cdot log_32=0\)
=>\(x\left(x+log_32\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x=-log_32\end{matrix}\right.\)
=>Chọn A
Câu 29:
\(3^{2x+8}-4\cdot3^{x+5}+27=0\)
=>\(3^{2x}\cdot3^8-4\cdot3^x\cdot3^5+27=0\)
=>\(\left(3^x\right)^2\cdot6561-972\cdot3^x+27=0\)
=>\(\left[{}\begin{matrix}3^x=\dfrac{1}{9}\\3^x=\dfrac{1}{27}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
=>Chọn A
. Chọn đáp án sai:
