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

1 Answer
Aug 29, 2016

Answer:

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.