1+log₆(4-x)≤log₆(16-x²)
1=log₆6¹=log₆6
log₆6+log₆(4-x) ≤log₆(16-x²)
log₆(6*(4-x))≤log₆(16-x²)
основание логарифма а=6, 6>1. знак неравенства не меняем
{24-6x≤16-x² {x²-6x+8≤0 {(x-2)*(x-4)≤0 (1)
4-x>0 -x>-4 x<4 (2)<br> 16-x²>0 (4-x)*(4+x)>0 (4-x)*(4+x)>0 (3)
+ - +
(1) -----------[2]-------[4]-------------->x
x∈[2;4]
\ \ \ \ \ \ \
(2) ----------(4)---------->x
x∈(-∞;4)
- + -
(3) ---------(-4)--------(4)---------->x
x∈(-4;4)
\ \ \ \ \ \ \ \ \ \
----------(-4)--------[2]------------([4])-------------->x
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
/ / / / / / / / / / / / / / / /
ответ: x∈[2;4)