Question

lg(4x+1)-Lg x=0 быстрее пожалуйста

Answer 1

ОДЗ :

$\displaystyle\bf\\\left \{ {{4x+1 > 0} \atop {x > 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x > -0,25} \atop {x > 0}} \right. \ \ \ \Rightarrow \ \ \ x > 0\\\\\\\lg(4x+1)-\ lgx=0\\\\\lg\frac{4x+1}{x} =0\\\\\\\frac{4x+1}{x} =1\\\\\\\frac{4x+1}{x} -1=0\\\\\\\frac{4x+1-x}{x} =0\\\\\\\left \{ {{3x+1=0} \atop {x\neq 0}} \right. \\\\\\3x+1=0\\\\3x=-1\\\\x=-\frac{1}{3} \\\\Otvet \ x\in \oslash$

Корней нет