Напишите решение с ОДЗ

Question 1

Question 2

$\displaystyle log_{1.5(x+1)}(3x+8)*log_{1.5(x+1)}\bigg( \frac{3x+8}{(1.5(x+1))^3}\bigg) \leq -2\\\\ODZ:$
$\left \{ {{3x+8\ \textgreater \ 0; \frac{3x+8}{(1.5(x+1))^3}\ \textgreater \ 0 } \atop {1,5(x+1)\ \textgreater \ 0;1.5(x+1) \neq 1};} \right.\\\\ \left \{ {{x\ \textgreater \ -8/3; x\ \textgreater \ -1} \atop {x \neq -1;x \neq - \frac{1}{3}}} \right. \\\\ODZ:x\ \textgreater \ -1; x \neq - \frac{1}{3}$

$\displaystyle log_{1.5(x+1)}(3x+8)*(log_{1.5(x+1)}(3x+8)-log_{1.5(x+1)}(1.5(x+1))^3)=\\\\=log_{1.5(x+1)}(3x+8)*(log_{1.5(x+1)}(3x+8)-3) \leq -2$
$\displaystyle log^2_{1.5(x+1)}(3x+8)-3log_{1.5(x+1)}(3x+8)+2 \leq 0\\\\log_{1.5(x+1)}(3x+8)=t\\\\t^2-3t+2 \leq 0\\\\1 \leq t \leq 2$

$\displaystyle 1 \leq log_{1.5(x+1)}(3x+8) \leq 2$

Два случая когда основание больше 1 и
когда основание больше 0 но меньше 1

1) Основание >1
$\displaystyle 1,5(x+1)\ \textgreater \ 1\\\\x\ \textgreater \ -1/3$

$\displaystyle \left \{ {{log_{1.5(x+1)}(3x+8) \geq 1} \atop {log_{1.5(x+1)}(3x+8) \leq 2}} \right.\\\\ \left \{ {{3x+8 \geq 1.5x+1.5} \atop {3x+8 \leq (1.5(x+1))^2}} \right.$

$\displaystyle 3x+8 \geq 1.5(x+1)\\\\1.5x \geq -6.5\\\\x \geq -13/3$
с учетом условия x>-1/3

$\displaystyle 3x+8 \leq 2.25(x^2+2x+1)|*4\\\\12x+32 \leq 9(x^2+2x+1)\\\\9x^2+6x-23 \geq 0\\\\(-oo; \frac{-1-2 \sqrt{6}}{3}][\frac{-1+2 \sqrt{6}}{3}; +oo)$
с учетом условия x>-1/3
$\displaystyle [\frac{-1+2 \sqrt{6}}{3}; +oo)$

2) основание больше 0 но меньше 1

$\displaystyle -1\ \textless \ x\ \textless \ -1/3\\\\ \left \{ {{log_{1.5(x+1)}(3x+8) \leq 2} \atop {log_{1.5(x+1)}(3x+8) \geq 1}} \right. \\\\ \left \{ {{3x+8 \geq (1.5(x+1))^2} \atop {3x+8 \leq 1.5(x+1)}} \right. \\\\$

$\displaystyle 3x+8 \leq 1.5(x+1)\\\\x \leq -13/3$
с учетом условия -1решений нет

ОТВЕТ:

$\displaystyle [\frac{-1+2 \sqrt{6}}{3};+oo)$