if b > or equal to 1 , Show that e^b > or equal to b^e?

Question

if b > or equal to 1 , Show that e^b > or equal to b^e I tried to differentiate but it didn't work

Answer 1

I'm assuming that e is the natural base e

take the log of both sides

ln(e^b) >= ln(b^e)
b >= e * ln(b)
e = 2.72
so you need to prove that b/ln(b) > 2.72

take the derivative of
d(x/ln(x)) = 0

1 - 1/ln(x) = 0
x = e (obvious since e^e = e^e, but we want to make sure that there is only a single minimum and not multiple ones)
take the second derivative to make sure that it is a minimum and not a maximum. I leave that as an exercise to you.

So every other point where b>=1
x/ln(x) is greater than 2.72.

Answer 2

based on the link below(dated 23 June 2007)

x^(1/x) is maximum when x = e

so b= e=> both are same

if b!=e then

b^(1/b) < e^(1/e)

raising to the power be(product of b and e) we get

b^e < e^b

Answer 3

i don't know