sangers1959

Verified answer

18.
4^x+12=7*2^x
2^x=t  ⇒
t²-7t+12=0   D=1
t₁=4   2^x=4  2^x=2^2   x₁=2
t₂=3   2^x=3  log₂2^x=log₂3   x₂=log₂3
Ответ:  х₁=2   х₂=log₂3.
19.
2^(2x-4)<cos(π/4)
2^(2x-4)<√2/2
2^(2x-4)<1/√2
2^(2x-4)<1/2^(1/2)
2^(2x-4)<2^(-1/2)
2x-4<-1/2
2x<3¹/₂=3,5
x<1,75.