Question 923463
The solution process and answer result are good.


A couple of formal differences could be made in place.


You would "take the log_base_e of both sides" at {{{e^(x^2)= e^(x+6)}}}
{{{ln(e^(x^2))=ln(e^(x+6))}}}
{{{(x^2)ln(e)=(x+6)*ln(e)}}}
{{{x^2=x+6}}}


The next step will give a more convenient form to work with,
{{{x^2=x+6}}}
{{{x^2-x-6=x+6-x-6}}}
{{{x^2-x-6=0}}}


Easy enough to factor,
{{{(x+2)(x-3)=0}}}
Which give the solutions you found.