(2) e^(1/x) > 1
When e^(1/x) = 1, (taking natural logs of both sides)
(1/x) = 0
x = infinity
Any positive number x that is less than infinity satisfies the inequality.
(1) The solution to
xln(x) + 1 = x + ln(x)
is x = 1. Values x between 0 and satisfy the inequality.
Solve for x?
xln(x) + 1 > x + ln(x)
e^1/x > 1
Any help would be appreciated. THANK YOU!
1 answer