Question

Solve `e^x-lnx<=e/x` ?

Solve for `x`,

`e^x-lnx<=e/x`

Answer 1

so the solution of this inequality make it true `x in(0.1]`

Explanation:

consider `f(x)=e^x-lnx-e/x` ,we have

`f'(x)=e^x-1/x+e/x^2`

argue that `f'(x)>0` for all real x and conclude noting that `f(1)=0`

`f(1)=e-ln1-e=0`

consider the limit of f as x goes to 0

`lim_(xrarr0)e^x-lnx-e/x`

`lim_(xrarr0^+)e^x-lnx-e/x=-oo`

In other words, by showing `f'(x)>0` you show that the function is strictly increasing, and if `f(1)=0` that means that `f(x)<0`
for `x<1` because the function always grows.

from the definition of `lnx`

`lnx` is defined for each `x>0`

from the definition of `e^x`

`e^x` is defined for each `x>=0`

but `e/x=e/0` undefined

so the solution of this inequality make it true `x in(0.1]`