Giải mỗi phương trình sau:

a) log₄ (x – 4) = –2;                  

b) log₃ (x² + 2x) = 1;

c) ${\log }_{25}\left(x^2-4\right)=\frac{1}{2};$    

d) log₉ [(2x – 1)²] = 2;

e) log(x² – 2x) = log(2x – 3);   

g) ${\log }_2\left(x^2\right)+{\log }_{\frac{1}{2}}\left(2x+8\right)=0.$ 

a) log₄ (x – 4) = –2 ⇔ x – 4 = 4^–2

^{$\iff x-4=\frac{1}{16}\iff x=\frac{65}{16}.$}

Vậy phương trình có nghiệm $x=\frac{65}{16}.$

b) log₃ (x² + 2x) = 1 ⇔ x² + 2x = 3¹

$\iff x^2+2x-3=0\iff \left[\begin{array}{l}x=-3\\ x=1\end{array}\right..$

Vậy phương trình có nghiệm x ∈ {– 3; 1}.

c) ${\log }_{25}\left(x^2-4\right)=\frac{1}{2}\iff x^2-4={25}^{\frac{1}{2}}$

$\iff x^2-4=5\iff x^2=9\iff \left[\begin{array}{l}x=3\\ x=-3\end{array}\right..$

Vậy phương trình có nghiệm x ∈ {– 3; 3}.

d) ${\log }_9\left[{\left(2x-1\right)}^2\right]=2\iff {\left(2x-1\right)}^2=9^2$

$\iff 4x^2-4x-80=0\iff \left[\begin{array}{l}x=-4\\ x=5\end{array}\right..$

Vậy phương trình có nghiệm x ∈ {– 4; 5}.

e) Ta có: $\log \left(x^2-2x\right)=\log \left(2x-3\right)$

 $\iff \left\{\begin{array}{l}{x}^2-2x=2x-3\\ 2x-3>0\end{array}\right.\iff \left\{\begin{array}{l}{x}^2-4x+3=0\\ x>\frac{3}{2}\end{array}\right.$

 $\iff \left\{\begin{array}{l}\left[\begin{array}{l}x=1\\ x=3\end{array}\right.\\ x>\frac{3}{2}\end{array}\right.\iff x=3.$

Vậy phương trình có nghiệm x = 3.

g) ${\log }_2\left(x^2\right)+{\log }_{\frac{1}{2}}\left(2x+8\right)=0.$

$\iff {\log }_2\left(x^2\right)+{\log }_{2^{-1}}\left(2x+8\right)=0$

⇔ log₂ (x²) – log₂ (2x + 8) = 0

⇔ log₂ (x²) = log₂ (2x + 8)

 $\iff \left\{\begin{array}{l}{x}^2=2x+8\\ 2x+8>0\end{array}\right.\iff \left\{\begin{array}{l}{x}^2-2x-8=0\\ x>-4\end{array}\right.$

$\iff \left\{\begin{array}{l}\left[\begin{array}{l}x=-2\\ x=4\end{array}\right.\\ x>-4\end{array}\right.\iff \left[\begin{array}{l}x=-2\\ x=4\end{array}\right.$

Vậy phương trình có nghiệm x ∈ {– 2; 4}.