Question

giải các phương trình sau

a) \(\log_5\left(4x-3\right)=2\)

b) \(\log_2x^2=2\)

c) \(\log_5\left(2x+1\right)=\log_5\left(-2x+3\right)\)

d) \(\ln\left(x^2-6x+7\right)=\ln\left(x-3\right)\)

e) \(\log\left(5x-1\right)=log\left(4-2x\right)\)

Answer 1

a: ĐKXĐ: \(4x-3>0\)

=>x>3/4

\(log_5\left(4x-3\right)=2\)

=>\(log_5\left(4x-3\right)=log_525\)

=>4x-3=25

=>4x=28

=>x=7(nhận)

b: ĐKXĐ: \(x\ne0\)

\(log_2x^2=2\)

=>\(log_2x^2=log_24\)

=>\(x^2=4\)

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{3}{2}\right\}\)

\(\log_52x+1=\log_5-2x+3\)

=>2x+1=-2x+3

=>4x=2

=>\(x=\dfrac{1}{2}\left(nhận\right)\)

d: ĐKXD: \(x\notin\left\{3\right\}\)

\(ln\left(x^2-6x+7\right)=ln\left(x-3\right)\)

=>\(x^2-6x+7=x-3\)

=>\(x^2-7x+10=0\)

=>(x-2)(x-5)=0

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\notin\left\{\dfrac{1}{5};2\right\}\)

\(log\left(5x-1\right)=log\left(4-2x\right)\)

=>5x-1=4-2x

=>7x=5

=>\(x=\dfrac{5}{7}\left(nhận\right)\)