Question 1172409
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Looking at base = e, and ln (the natural log)...

A graph showing {{{y = e^x}}}, {{{y = x}}}, and {{{ y = ln(x) }}} is a quick way to see they will never intersect.


Aside from that... for {{{ x<0}}}, clearly {{{e^x > x }}} as {{{ e^x >0 }}} while {{{x<0}}}.  For the half-plane x>=0, one can write the Taylor series expansion for {{{e^x}}}:  
  {{{e^x}}} = {{{sum((x^n/n!), n=0, infinity)}}}

Pull out the first two terms from the RHS:
  {{{e^x}}} = {{{1 + x + sum((x^n/n!), n=2, infinity)}}}

So clearly {{{e^x > x }}} for {{{ x>= 0}}} and combined with the first part, {{{e^x > x }}} for all x.


To prove  {{{ ln(x) }}} <  {{{x }}}  (here the domain of x is x>0, as ln(x) is not defined for {{{x<=0}}} ),  we can use the result from above as the starting point:

       {{{ e^x  >  x}}}
Take ln() both sides:
       {{{ ln(e^x) }}} > {{{ ln(x) }}}
       {{{  x }}} > {{{ ln(x) }}}   , x>0