Question

lg(3x^2+12x+19)-lg(3x+4)=1

Answer 1

$\lg(3x^2+12x+19)-\lg(3x+4)=1$

одз:

$\left \{ {{3x^2+12x+19>0} \atop {3x+4>0}} \right. \\3x^2+12x+19=0\\\left \{ {{D=12^2-12*19<0} \atop {3>0}} \right. \Rightarrow 3x^2+12x+19>0 ,\forall \ x\in R\\3x+4>0\\3x>-4\\x>-\frac{4}{3}\\x\in(-\frac{4}{3};+\infty)\\\left \{ {{x\in R} \atop {x\in(-\frac{4}{3};+\infty)}} \right. \Rightarrow x\in(-\frac{4}{3};+\infty)$

решаем уравнение:

$\lg(3x^2+12x+19)-\lg(3x+4)=1\\\lg(\frac{3x^2+12x+19}{3x+4})=1\\\frac{3x^2+12x+19}{3x+4}=10^1\\3x^2+12x+19=10(3x+4)\\3x^2+12x+19=30x+40\\3x^2-18x-21=0\\x^2-6x-7=0\\D=36+28=64=8^2\\x_1=\frac{6-8}{2}=-1\in (-\frac{4}{3};+\infty)\\x_2=\frac{6+8}{2}=7\in (-\frac{4}{3};+\infty)$

Ответ: -1; 7