Question 737740
{{{ln(1+x^2)=1+2lnx}}}

convert 1 to {{{ln(e^1)=lne}}} and {{{2lnx=ln(x^2)}}}

so we have:

{{{ln(1+x^2)=lne+ln(x^2)=ln(ex^2)}}}

so {{{1+x^2=ex^2}}} 
{{{1=ex^2-x^2=x^2(e-1)}}}
{{{x^2= 1/(e-1)}}}
{{{x=1/(sqrt(e-1))=0.76287=0.763}}}