Giải phương trình sau: logx + log(x – 3) = 1.

logx + log(x – 3) = 1

Điều kiện xác định là $\left\{\begin{array}{l}x>0\\ x-3>0\end{array}\right.,$ tức là x > 3. Ta có:

logx + log(x – 3) = 1

$$\begin{gathered} \iff \left\{\begin{array}{l}x>3\\ log\left[x\left(x–3\right)\right]=1\end{array}\right.\iff \left\{\begin{array}{l}x>3\\ x\left(x–3\right)={10}^1\end{array}\right. \\ \iff \left\{\begin{array}{l}x>3\\ x^2-3x-10=0\end{array}\right.\iff \left\{\begin{array}{l}x>3\\ \left(x-5\right)\left(x+2\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}x>3\\ \left[\begin{array}{l}x-5=0\\ x+2=0\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}x>3\\ \left[\begin{array}{l}x=5\\ x=-2\end{array}\right.\end{array}\right.\iff x=5 \end{gathered}$$

Vậy phương trình có nghiệm là x = 5.