Question

Bài 1: Giải phương trình:

a) log₃(2x+1)-log₃(x-1)=1

b) log₂(x-1)+log₂(x-2)=log₅(125)

c) log₂(sinx)+log₂(cosx)=-2, x thuộc (0;2π)

d) log_√2(x+1)=log₂(x²+2)-1

e) 3log₃(x-1)-log_1/5(x-5)³=3

f) log(x²-x-1)=log(2x+1)

Bài 2: Giải phương trình:

a) log²_1/3x-5log₃x+4=0

b) log²₂(4x)-log_√2(2x)=5

Bài 3: Tìm m để phương trình:

a) log_1/3(x+m)+log₃(2-x)=0 có nghiệm

b) log²₂x-7log₂x+m-3=0 có 2 nghiệm x₁,x₂ thỏa mãn x₁.x₂=16

c) log²₃(3x)+log₃x+m-1=0 có 2 nghiệm phân biệt thuộc (0;1)

Answer 1

Bài 2:

a: \(log^2_{\dfrac{1}{3}}x-5\cdot log_3x+4=0\)

=>\(log_{\dfrac{1}{3}}^2x+5\cdot log_{\dfrac{1}{3}}x+4=0\)

=>\(\left(log_{\dfrac{1}{3}}x+1\right)\left(log_{\dfrac{1}{3}}x+4\right)=0\)

=>\(\left[{}\begin{matrix}log_{\dfrac{1}{3}}x=-1\\log_{\dfrac{1}{3}}x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=81\end{matrix}\right.\)

b: \(log_2^24x-log_{\sqrt{2}}2x=5\)

=>\(log_2^24x-2\cdot log_22x=5\)

=>\(\left(log_24x\right)^2-2\cdot log_22x=5\)

=>\(\left(1+log_22x\right)^2-2\cdot log_22x=5\)

=>\(\left(log_22x\right)^2+1=5\)

=>\(\left(log_22x\right)^2=4\)

=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{8}\end{matrix}\right.\)

Bài 1:

a:

ĐKXĐ: x>1

 \(log_3\left(2x+1\right)-log_3\left(x-1\right)=1\)

=>\(log_3\left(\dfrac{2x+1}{x-1}\right)=1\)

=>\(\dfrac{2x+1}{x-1}=3\)

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

=>3x-3=2x+1

=>x=4(nhận)

b:

ĐKXĐ: x>2

 \(log_2\left(x-1\right)+log_2\left(x-2\right)=log_5\left(125\right)\)

=>\(log_2\left[\left(x-1\right)\left(x-2\right)\right]=3\)

=>\(\left(x-1\right)\left(x-2\right)=2^3=8\)

=>\(x^2-3x-6=0\)

=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{2}\left(nhận\right)\\x=\dfrac{3-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)

c: \(log_2\left(sinx\right)+log_2\left(cosx\right)=-2\)

=>\(log_2\left(sinx\cdot cosx\right)=-2\)

=>\(log_2\left(\dfrac{1}{2}\cdot sin2x\right)=-2\)

=>\(\dfrac{1}{2}\cdot sin2x=\dfrac{1}{4}\)

=>\(sin2x=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x=\dfrac{\Omega}{6}+k2\Omega\\2x=\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{12}+k\Omega\\x=\dfrac{5}{12}\Omega+k\Omega\end{matrix}\right.\)

\(x\in\left(0;2\Omega\right)\)

=>\(\left[{}\begin{matrix}\dfrac{\Omega}{12}+k\Omega\in\left(0;2\Omega\right)\\\dfrac{5}{12}\Omega+k\Omega\in\left(0;2\Omega\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k+\dfrac{1}{12}\in\left(0;2\right)\\k+\dfrac{5}{12}\in\left(0;2\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}k\in\left(-\dfrac{1}{12};\dfrac{23}{12}\right)\\k\in\left(-\dfrac{5}{12};\dfrac{19}{12}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left(0;1\right)\\k\in\left(0;1\right)\end{matrix}\right.\)

=>\(x\in\left\{\dfrac{\Omega}{12};\dfrac{13}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{17}{12}\Omega\right\}\)