Question

Решить неравенство 9log12(x^2-3x-4)≤10+log12((x+1)^9/(x-4))

Answer 1

$9log _{12}( x^{2} -3x-4) \leq 10+log _{12} \frac{(x+1) ^{9} }{x-4}$
Найдём ОДЗ:
$\left \{ {{ x^{2} -3x-4\ \textgreater \ 0} \atop { \frac{(x+1) ^{9} }{x-4}\ \textgreater \ 0 }} \right. \\\\(x-4)(x+1)\ \textgreater \ 0$
x ∈ (- ∞ ; - 1) ∪ (4 ; + ∞)

$log _{12}[(x-4)(x+1)] ^{9}-log _{12} \frac{(x+1) ^{9} }{x-4} \leq 10\\\\log _{12} \frac{(x-4) ^{9}(x+1) ^{9}(x-4) }{(x+1) ^{9} } \leq 10\\\\log _{12}(x-4) ^{10} \leq 10 \\\\10log _{12}(x-4) \leq 10\\\\log _{12}|x-4| \leq 1\\\\|x-4|-12 \leq 0\\\\(x-4-12)(x-4+12) \leq 0\\\\(x-16)(x-8) \leq 0$
x ∈ [- 8 ; 16]
С учётом ОДЗ , окончательный ответ:
x ∈ [- 8 ; - 1) ∪ (4 ; 16]