Question

How do you solve `e^x - lnx= 0`?

Answer 1

Not real solution.

Explanation:

`e^x - log_e x= 0`

Considering `x > 0` for the feasibility of `log_e x` we have

`e^x-log_e x equiv e^(e^x)= x`

but

`e^(e^x) = 1 +e^x+1/(2!)e^(2x)+ cdots`
`=1 + (1+x+x^2/(2!)+ cdots)+1/(2!)e^(2x)+ cdots > x`

so `e^x - log_e x= 0` has not real solution.