Question

Giải các phương trình sau:

a) \({\left( {\frac{1}{4}} \right)^{x - 2}} = \sqrt 8 \);

b) \({9^{2x - 1}} = {81.27^x}\);

c) \(2{\log _5}\left( {x - 2} \right) = {\log _5}9\);

d) \({\log _2}\left( {3{\rm{x}} + 1} \right) = 2 - {\log _2}\left( {x - 1} \right)\).

Answer 1

\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)

\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)

Answer 2

c, ĐK: \(x-2>0\Rightarrow x>2\)

\(2log_5\left(x-2\right)=log_59\\ \Leftrightarrow log_5\left(x-2\right)^2=log_59\\ \Leftrightarrow\left(x-2\right)^2=3^2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.

d, ĐK: \(x-1>0\Leftrightarrow x>1\)

\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)