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a) \({3^{{x^2} - 4x + 5}} = 9 \Leftrightarrow {x^2} - 4x + 5 = 2 \Leftrightarrow {x^2} - 4x + 3 = 0 \Leftrightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = 1\end{array} \right.\)
Vậy phương trình có nghiệm là \(x \in \left\{ {1;3} \right\}\)
b) \(0,{5^{2x - 4}} = 4 \Leftrightarrow 2x - 4 = {\log _{0,5}}4 \Leftrightarrow 2x = 2 \Leftrightarrow x = 1\)
Vậy phương trình có nghiệm là x = 1
c) \({\log _3}(2x - 1) = 3\) ĐK: \(2x - 1 > 0 \Leftrightarrow x > \frac{1}{2}\)
\( \Leftrightarrow 2x - 1 = 27 \Leftrightarrow x = 14\) (TMĐK)
Vậy phương trình có nghiệm là x = 14
d) \(\log x + \log (x - 3) = 1\) ĐK: \(x - 3 > 0 \Leftrightarrow x > 3\)
\(\begin{array}{l} \Leftrightarrow \log \left( {x.\left( {x - 3} \right)} \right) = 1\\ \Leftrightarrow {x^2} - 3x = 10\\ \Leftrightarrow {x^2} - 3x - 10 = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = - 2 (loại) \,\,\,\\x = 5 (TMĐK) \,\,\,\,\,\,\,\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm x = 5
\(a,\left(0,3\right)^{x-3}=1\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\\ b,5^{3x-2}=25\\ \Leftrightarrow3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\\ c,9^{x-2}=243^{x+1}\\ \Leftrightarrow3^{2x-4}=3^{5x+5}\\ \Leftrightarrow2x-4=5x+5\\ \Leftrightarrow3x=-9\\ \Leftrightarrow x=-3\)
d, Điều kiện: \(x>-1;x\ne0\)
\(log_{\dfrac{1}{x}}\left(x+1\right)=-3\\ \Leftrightarrow x+1=x^3\\ x\simeq1,325\left(tm\right)\)
e, Điều kiện: \(x>\dfrac{5}{3}\)
\(log_5\left(3x-5\right)=log_5\left(2x+1\right)\\ \Leftrightarrow3x-5=2x+1\\ \Leftrightarrow x=6\left(tm\right)\)
f, Điều kiện: \(x>\dfrac{1}{2}\)
\(log_{\dfrac{1}{7}}\left(x+9\right)=log_{\dfrac{1}{7}}\left(2x-1\right)\\ \Leftrightarrow x+9=2x-1\\ \Leftrightarrow x=10\left(tm\right)\)
\(a,3^{x-1}=27\\ \Leftrightarrow3^{x-1}=3^3\\ \Leftrightarrow x-1=3\\ \Leftrightarrow x=4\\ b,100^{2x^2-3}=0,1^{2x^2-18}\\ \Leftrightarrow10^{4x^2-6}=10^{-2x^2+18}\\ \Leftrightarrow4x^2-6=-2x^2+18\\ \Leftrightarrow6x^2=24\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)
\(c,\sqrt{3}e^{3x}=1\\ \Leftrightarrow e^{3x}=\dfrac{1}{\sqrt{3}}\\ \Leftrightarrow3x=ln\left(\dfrac{1}{\sqrt{3}}\right)\\ \Leftrightarrow x=\dfrac{1}{3}ln\left(\dfrac{1}{\sqrt{3}}\right)\)
\(d,5^x=3^{2x-1}\\ \Leftrightarrow2x-1=log_35^x\\ \Leftrightarrow2x-1-xlog_35=0\\ \Leftrightarrow x\left(2-log_35\right)=1\\ \Leftrightarrow x=\dfrac{1}{2-log_35}\)
\(a,2^{3x-1}=2^{-\left(x+1\right)}\Rightarrow3x-1=-\left(x+1\right)\Rightarrow x=\dfrac{1}{2}\)
\(b,ln\left(2e^{2x}\right)=ln5\)
\(\Rightarrow ln2+lne^{2x}=ln5\)
\(\Rightarrow ln2+2x=ln5\)
\(\Rightarrow2x=ln5-ln2=ln\dfrac{5}{2}\)
Như vậy \(x=\dfrac{1}{2}ln\dfrac{5}{2}\)
a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)
=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)
=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)
b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)
=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)
c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)
=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)
=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)
=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)
=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)
=>\(x=\dfrac{1}{4}\Omega+k\Omega\)
\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)
\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)
c, ĐK: \(x>-7\)
\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)
Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)
d, ĐK: \(x>\dfrac{1}{2}\)
\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)
Kết hợp với ĐKXĐ, ta được: \(x\ge8\)
\(a,3^{1-2x}=4^x\\ \Leftrightarrow1-2x=log_34^x\\ \Leftrightarrow1-2x=xlog_34\\ \Leftrightarrow2x+xlog_34=1\\ \Leftrightarrow x\left(2+log_34\right)=1\\ \Leftrightarrow x=\dfrac{1}{2+log_34}=\dfrac{1}{log_39+log_34}=\dfrac{1}{log_336}=log_{36}3\)
b, ĐK: \(x>-1\)
\(log_3\left(x+1\right)+log_3\left(x+4\right)=2\\ \Leftrightarrow log_3\left(x^2+5x+4\right)=2\\ \Leftrightarrow x^2+5x+4=9\\ \Leftrightarrow x^2+5x-5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+3\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{-5-3\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)
\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)
\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)
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) \(0,{1^{2x - 1}} \le 0,{1^{2 - x}} \Leftrightarrow 2x - 1 \ge 2 - x \Leftrightarrow 3x \ge 3 \Leftrightarrow x \ge 1\)
b) \({3.2^{x + 1}} \le 1 \Leftrightarrow {2^{x + 1}} \le \frac{1}{3} \Leftrightarrow x + 1 \le {\log _2}\frac{1}{3} \Leftrightarrow x \le - {\log _2}3 - 1 = - {\log _2}3 - {\log _2}2 = - {\log _2}6\)
\(a,5^{2x-1}=25\\ \Leftrightarrow5^{2x-1}=5^2\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\)
\(b,3^{x+1}=9^{2x+1}\\ \Leftrightarrow3^{x+1}=3^{4x+2}\\ \Leftrightarrow x+1=4x+2\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\)
\(c,10^{1-2x}=100000\\ \Leftrightarrow10^{1-2x}=10^5\\ \Leftrightarrow1-2x=5\\ \Leftrightarrow2x=-4\\ \Leftrightarrow x=-2\)