Question

решите неравенство 0,5<2^1-x<32

Answer 1

$0,5 textless 2^{1-x} textless 32\\ frac{1}{2} textless 2^{1-x} textless 2^5\\2^{-1} textless 2^{1-x} textless 2^5; ; Rightarrow ; ; ; -1 textless 1-x textless 5\\-2 textless -x textless 4; |cdot (-1)\\-4 textless x textless 2\\xin (-4,2)$

Answer 2

0,5<2^(1-х)<32
2(-1)<2^(1-х)<2^5
-1<(1-х)<5
-1<-(х-1)<5
-5<х-1<1
-4<х<2