Question

关于x的不等式：8^x+2^x-2

Answer 1

令2^x=t,显然,有t>0

∴8^x+2^x-2=t^3+t-2=(t^3-1)+(t-1)=f(t)

∵t^3-1=(t-1)(t^2+t+1)

∴f(t)=(t^3-1)+(t-1)=(t-1)(t^2+t+2)

8^x+2^x-2＜0即f(t)=(t^3-1)+(t-1)=(t-1)(t^2+t+2)＜0

∵t^2+t+2＞0恒成立,解得t-1＜0,即t＜1

∴2^x＜1

∴x＜0