Question 178937
{{{ln(9-x^2)=2}}}
{{{(9-x^2)=e^2}}}
{{{x^2=9-e^2}}}
{{{x=0 +- sqrt(9-e^2)}}} 
or approximately,
{{{x=0 +- sqrt(9-7.39)}}} 
{{{x=0 +- sqrt(1.61)}}} 
{{{x=0 +- 1.27}}}