Упражнение

[Гост]

log 3 (2x -1) + log 3 (x - 4) = 2

[ammornil]

log 3 (2x -1) + log 3 (x - 4) = 2

[tex]log_{3}(2x-1)+log_{3}(x-4)=2[/tex]

DM: [tex]\begin{array}{|l} 2x-1>0 \\ x-4>0 \end{array} \Rightarrow \begin{array}{|l} 2x>1 \\ x>4 \end{array} \Rightarrow \begin{array}{|l} x>{\Large \frac{1}{2} } \\x>4 \end{array} \Rightarrow x \in (4;+ \infty)[/tex]

[tex]log_{3}[(2x-1)(x-4)]=log_{3}(3^{2}) \Leftrightarrow (2x-1)(x-4)=9 \Leftrightarrow 2x^{2}-4.2x-x+4-9=0 \Leftrightarrow 2x^{2}-9x-5=0[/tex]
[tex]D=(-9)^{2}-4.2.(-5)=81+40=121[/tex]
[tex]x_{_{1,2}}=\frac{9\pm11}{2.2} \Rightarrow \begin{cases} x_{_{1}}={\Large-\frac{1}{2} } \notin DM \\ x_{_{2}}=5 \in DM \end{cases}[/tex]

[tex][/tex]