Equation with log LN

[parisse]

You don't need multi-precision floats, but you have to desingularize the equation if you want to improve the accuracy of 1+exp(-99)/2. The next command will also work on the Prime:
f:=x^2-100-log(x^2-1); x:=1+h; h:=exp(-99)/2*(1+k); [kk]:=fsolve(normal(f),k=-0.1..0.1);
Therefore x-1 is approximatively exp(-99)/2*(1+kk)
kk being approx 4.7e-13, in fact 1+exp(-99)/2 is already a numeric approximation of x with 50 digits.