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tham khảo
a)Điều kiện \(3-x>0\) hay \(x< 3\)
\(4-log\left(3-x\right)=3log\left(3-x\right)=1\Leftrightarrow10^1=3-x\)
Vậy nghiệm của phương trình là \(x=2\) thỏa mãn điều kiện
b) Điều kiện \(x+2>0\) và \(x-1>0\) tức là \(x>1\)
\(\left(x+2\right)\left(x-1\right)=2\Rightarrow x^2+x-4=0\)
Vậy pt có nghiệm \(x=\dfrac{-1+\sqrt{17}}{2}\)
a) \({\log _{\frac{1}{2}}}\left( {x - 2} \right) = - 2\)
Điều kiện: \(x - 2 > 0 \Leftrightarrow x > 2\)
Vậy phương trình có nghiệm là \(x = 6\).
b) \({\log _2}\left( {x + 6} \right) = {\log _2}\left( {x + 1} \right) + 1\)
Điều kiện: \(\left\{ \begin{array}{l}x + 6 > 0\\x + 1 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > - 6\\x > - 1\end{array} \right. \Leftrightarrow x > - 1\)
Vậy phương trình có nghiệm là \(x = 4\).
\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)
\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)
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 \(x=\dfrac{5}{3}\)
a, ĐK: \(x-2>0\Rightarrow x>2\)
\(log_2\left(x-2\right)< 2\\ \Leftrightarrow x-2< 4\\ \Leftrightarrow x< 6\)
Kết hợp với ĐKXĐ, ta được: \(2< x< 6\)
b, ĐK: \(2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)
\(log\left(x+1\right)\ge log\left(2x-1\right)\\ \Leftrightarrow x+1\ge2x-1\\ \Leftrightarrow x\le2\)
Kết hợp với ĐKXĐ, ta được: \(\dfrac{1}{2}< x\le2\)
a, ĐK: \(x+1>0\Leftrightarrow x>-1\)
\(log\left(x+1\right)=2\\ \Leftrightarrow x+1=10^2\\ \Leftrightarrow x+1=100\\ \Leftrightarrow x=99\left(tm\right)\)
b, ĐK: \(\left\{{}\begin{matrix}x-3>0\\x>0\end{matrix}\right.\Rightarrow x>3\)
\(2log_4x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2\left(x^2-3x\right)=2\\ \Leftrightarrow x^2-3x=4\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
c, ĐK: \(x>1\)
\(lnx+ln\left(x-1\right)=ln4x\\ \Leftrightarrow ln\left[x\left(x-1\right)\right]-ln4x=0\\ \Leftrightarrow ln\left(\dfrac{x-1}{4}\right)=0\\ \Leftrightarrow\dfrac{x-1}{4}=1\\ \Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\)
d, ĐK: \(\left\{{}\begin{matrix}x^2-3x+2>0\\2x-4>0\end{matrix}\right.\Rightarrow x>2\)
\(log_3\left(x^2-3x+2\right)=log_3\left(2x-4\right)\\ \Leftrightarrow x^2-3x+2=2x-4\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
a)
ĐK: \(\left\{{}\begin{matrix}2x-4>0\\x-1>0\end{matrix}\right.\Leftrightarrow x>1\)
\(\log_5\left(2x-4\right)+\log_{\dfrac{1}{5}}\left(x-1\right)=0\\ \Leftrightarrow\log_5\left(2x-4\right)-\log_5\left(x-1\right)=0\\ \Leftrightarrow\log_5\left(\dfrac{2x-4}{x-1}\right)=\log_51\\ \Leftrightarrow\dfrac{2x-4}{x-1}=1\\ \Leftrightarrow2x-4=x-1\\ \Leftrightarrow x=3\left(tm\right)\)
Vậy x = 3.
b) ĐK: x > 0
\(\log_2x+\log_4x=3\\ \Leftrightarrow\log_2x+\dfrac{1}{2}\log_2x=3\\ \Leftrightarrow\left(1+\dfrac{1}{2}\right)\log_2x=3\\ \Leftrightarrow\dfrac{3}{2}\log_2x=3\\ \Leftrightarrow\log_2x=2\\ \Leftrightarrow x=4\left(tm\right)\)
Vậy x= 4
a: \(log_2\left(M\cdot N\right)=log_2\left(2^5\cdot2^3\right)=log_2\left(2^8\right)=8\)
\(log_2M+log_2N=log_22^5+log_22^3=5+3=8\)
=>\(log_2\left(MN\right)=log_2M+log_2N\)
b: \(log_2\left(\dfrac{M}{N}\right)=log_2\left(\dfrac{2^5}{2^3}\right)=log_2\left(2^2\right)=2\)
\(log_2M-log_2N=log_22^5-log_22^3=5-3=2\)
=>\(log_2\left(\dfrac{M}{N}\right)=log_2M-log_2N\)
a) \(log_2\left(3-2x\right)\) xác định khi \(3-2x>0\) hay \(x< \dfrac{3}{2}\)
b) \(log_3\left(x^2+4x\right)\) xác định khi \(x^2+4x>0\) hay \(x>0\) hoặc \(x< -4\)
a: \(log_2\left(mn\right)=log_2\left(2^7\cdot2^3\right)=7+3=10\)
\(log_2m+log_2n=log_22^7+log_22^3=7+3=10\)
=>\(log_2\left(mn\right)=log_2m+log_2n\)
b: \(log_2\left(\dfrac{m}{n}\right)=log_2\left(\dfrac{2^7}{2^3}\right)=7-3=4\)
\(log_2m-log_2n=log_22^7-log_22^3=7-3=4\)
=>\(log_2\left(\dfrac{m}{n}\right)=log_2m-log_2n\)
a) \(\log_2\left(mn\right)=\log_2\left(2^7.2^3\right)=\log_22^{7+3}=\log_22^{10}=10.\log_22=10.1=10\)
\(\log_2m+\log_2n=\log_22^7+\log_22^3=7\log_22+3\log_22=7.1+3.1=7+3=10\)
b) \(\log_2\left(\dfrac{m}{n}\right)=\log_2\dfrac{2^7}{2^3}=\log_22^4=4.\log_22=4.1=4\)
\(\log_2m-\log_2n=\log_22^7-\log_22^3=7.\log_22-3\log_22=7.1-3.1=4\)
\(A=log_2\left(x^3-x\right)-log_2\left(x+1\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x^3-x}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)=log_2x\)