Giải các phương trình logarit sau:

1) $3^{2x^2-x}=3$

2) $2^{x-1}.3^x=18$

3) $2^{x-6}={\left(\frac{1}{4}\right)}^{3x+1}$

4) ${\left(\frac{1}{25}\right)}^{x+1}=125^{2x}$

5) ${\left(\frac{2020}{2021}\right)}^{4x}={\left(\frac{2021}{2020}\right)}^{2x-6}$

6) $2^{\sqrt{x+3}}=2^{3-x}$

7) ${\log }_3\left(x^2+4x-3\right)={\log }_3\left(x+1\right)$

8) $2{\log }_2\left(2x+3\right)={\log }_2{x}^2$

9) ${\log }_{2}x+{\log }_2\left(x-3\right)=2$

10) ${\log }_{5}x={\log }_5\left(x+6\right)-{\log }_5\left(x+2\right)$

1) Ta có $3^{2x^2-x}=3\iff 2x^2-x=1\iff 2x^2-x-1=0\iff \left[\begin{array}{c}x=1\\ x=-\frac{1}{2}\end{array}\right].$

2) Ta có $2^{x-1}.3^x=18\iff \frac{2^x}{2}.3^x=18\iff 6^x=36\iff x=2$.

3) Ta có: $2^{x-6}={\left(\frac{1}{4}\right)}^{3x+1}\iff 2^{x-6}=2^{-2(3x+1)}\iff x-6=-6x-2\iff 7x=4\iff x=\frac{4}{7}$

4) Phương trình đã cho ${\left(\frac{1}{25}\right)}^{x+1}=125^{2x}\iff 5^{-2(x+1)}=5^{6x}\iff -2(x+1)=6x\iff 8x=-2\iff x=-\frac{1}{4}$.

5) Ta có ${\left(\frac{2020}{2021}\right)}^{4x}={\left(\frac{2021}{2020}\right)}^{2x-6}\iff {\left(\frac{2020}{2021}\right)}^{4x}={\left(\frac{2020}{2021}\right)}^{-2x+6}\iff 4x=-2x+6\iff x=1$.

6) Điều kiện $x+3\ge 0\iff x\ge -3$

Ta có: $2^{\sqrt{x+3}}=2^{3-x}\iff \sqrt{x+3}=3-x\iff \{\begin{array}{l}3-x\ge 0\\ x+3=9-6x+x^2\end{array}\iff \{\begin{array}{l}x\le 3\\ x^2-7x+6=0\end{array}\iff \{\begin{array}{l}x\le 3\\ \left[\begin{array}{c}x=1\\ x=6\end{array}\right]\end{array}$

7) Ta có ${\log }_3\left(x^2+4x-3\right)={\log }_3\left(x+1\right)\iff \{\begin{array}{l}{x}^2+4x-3=x+1\\ x>-1\end{array}\iff \{\begin{array}{l}{x}^2+3x-4=0\\ x>-1\end{array}\iff x=1$.

8) Ta có: $2{\log }_2\left(2x+3\right)={\log }_2{x}^2\iff \{\begin{array}{l}2x+3>0\\ x^2>0\\ {\log }_2{\left(2x+3\right)}^2={\log }_2{x}^2\end{array}\iff \{\begin{array}{l}2x+3>0\\ {\left(2x+3\right)}^2=x^2\end{array}$

$\iff \{\begin{array}{l}x>-\frac{3}{2}\\ \left[\begin{array}{c}2x+3=x\\ 2x+3=-x\end{array}\right]\end{array}\iff \{\begin{array}{l}x>-\frac{3}{2}\\ \left[\begin{array}{c}x=-3\\ x=-1\end{array}\right]\end{array}\iff \left[\begin{array}{c}x=-3\\ x=-1\end{array}\right]$. Vậy tập nghiệm của phương trình là $\left\{-3;-1\right\}$.

9) Điều kiện: $\{\begin{array}{l}x>0\\ x-3>0\end{array}\Rightarrow x>3$.

Ta có: ${\log }_{2}x+{\log }_2(x-3)=2\iff {\log }_2(x(x-3))=2\iff x^2-3x-4=0\iff \left[\begin{array}{cc}x=-1& (loại)\\ x=4& (tm)\end{array}\right]$

Vậy phương trình có nghiệm duy nhất $x=4$.

10) Điều kiện $\{\begin{array}{l}x>0\\ x+6>0\\ x+2>0\end{array}\iff \{\begin{array}{l}x>0\\ x>-6\\ x>-2\end{array}\iff x>0$.

Ta có: ${\log }_{5}x={\log }_5(x+6)-{\log }_5(x+2)\iff {\log }_{5}x+{\log }_5(x+2)={\log }_5(x+6)$

$\iff {\log }_5[x(x+2)]={\log }_5(x+6)\iff x(x+2)=x+6\iff \left[\begin{array}{c}x=2\\ x=-3\end{array}\right]$

So với điều kiện, suy ra $x=2$. Vậy $S=\{2\}$.