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

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Answer 1 · 2016-08-29T15:16:30

Not real solution.

$e^x - log_e x= 0$

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.

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