Найдите:

$y=ln(x+8)+ln(x+7)-1.5x-3\\\\y'=\frac{1}{x+8}+\frac{1}{x+7}-1.5 \leq 0\\\\\frac{x+7+x+8-1.5(x+7)(x+8)}{(x+7)(x+8)} \leq 0\\\\\frac{2x+15-1.5(x^2+7x+8x+56)}{(x+8)(x+7)} \leq 0\\\\\frac{4x+30-3(x^2+15x+56)}{(x+7)(x+8)} \leq 0\\\\\frac{4x+30-3x^2-45x-168}{(x+7)(x+8)} \leq 0\\\\\frac{-3x^2-41x-138}{(x+7)(x+8)} \leq 0\\\\\frac{3x^2+41x+138}{(x+7)(x+8)} \geq 0$

$D=41^2-4*3*138=25\\\\x_1=\frac{-41+5}6=-\frac{36}6=-6\\\\x_2=\frac{-41-6}6=-\frac{46}6=-\frac{23}3\\\\\frac{(x+6)(3x+23)}{x+7)(x+8)} \geq 0$

Метод интервалов:

$.....+....-8.....-.....-\frac{23}3.....+.....-7.......-......-6......+.......$

$x\in(-\infty;-8)U[-\frac{23}3;-7)U[-6;+\infty)$