One way is Newton's method.
It simplifies to:
\(\displaystyle f(x)=xln(x)-11.665293x+0.4\)
\(\displaystyle f'(x)=ln(x)-10.6652935059\)
\(\displaystyle x_{n+1}=x-\frac{xln(x)-11.665293x+0.4}{ln(x)-10.6652935059}\)
Try an initial guess of, say, 0.5
Several iterations gives:
\(\displaystyle .0261265....\)
There is another solution as well. Much larger. Try finding it by using a much larger initial guess.
