Question

Помогите, пожалуйста, решить логарифмическое уравнение

Answer 1

ОДЗ :

$\displaystyle\bf\\\left\{\begin{array}{ccc}3x^{2} +x-5 > 0\\x+2 > 0\\x+2\neq 1\end{array}\right \ \ \ \Rightarrow \ \ \ \left\{\begin{array}{ccc}3x^{2} +x-5 > 0\\x > -2\\x\neq -1\end{array}\right\\\\\\\log_{x+2} \Big(3x^{2} +x-5\Big)=2\\\\\Big(x+2\Big)^{2} =3x^{2}+x-5\\\\x^{2} +4x+4=3x^{2} +x-5\\\\2x^{2}-3x-9=0\\\\D=(-3)^{2} -4\cdot 2\cdot(-9)=9+72=81=9^{2} \\\\\\x_{1} =\frac{3+9}{4} =3\\\\\\x_{2} =\frac{3-9}{4} =-1,5$

Проверка :

$\displaystyle\bf\\3x^{2} +x-5 > 0\\\\1) \ x=3\\\\3\cdot 3^{2} +3-5=3\cdot 9-2=27-2=25 > 0\\\\2) \ x=-1,5\\\\3\cdot(-1,5)^{2} +(-1,5)-5=3\cdot 2,25-1,5-5=6,75-6,5=0,25 > 0\\\\Otvet : \ 3 \ ;\ -1,5$