Log2 16x≥(2log2 16x) /log2 0.5x
log2 16 +log2 x ≥ 2*(log2 16 +log2 x) / (log2 0.5 +log2 x)
4+log2 x≥ 2*(4+log2 x) / (-1+log2 x)
(4+log2 x )(log2 x -1)- 2*(log2 x+4) ) /(log2 x-1)≥0
(4+log2 x) ( log2 x -1 -2) /log2 x -1 ≥0
( 4+log2 x) ( log2 x -3)/ log2 x -1≥0
найдем нули функции 4+log2 x=0 log2 x=-4 x=1/16
log2 x - 3=0 x=8. log2 x=1 x=2
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0 - 1/16 + 2 - 8 +
x∈[1/16. 2) ∪ [8. ∞)
может так?