Its getting late here so I may be incorrect, but we could use the Taylor Series Expansion for The Natural Logarithm, as Tom mentioned. It would also be an easier calculation if you used the other expression,
\ln (x+1)=\sum_{n=0}^{infinity} \frac {-1^n \cdot x^{n+1}}{n+1}.
We could truncate the expression at the degree to which you can solve. It is possible to solve the polynomials roots analytically for up to degree 5, but I think you will find it gets difficult after Quadratics (a substitution will not make it reducible to a quadratic).
To be even more correct, you would express the rest of the series in the form of a Lagrange/Cauchy Remainder term.
We could also use a a variety of fixed-point iteration methods, Such as halving the interval, Newtons method, Halleys method (similar to Newtons but with conditional cubic convergence), or the runge-kutta method.
For an extremely advanced method, use halving the interval until you have the root of f(x)=ln x + 0.5x^2 over a small closed interval. The interval must be small enough so that its Lipschitz constant, or maximum slope between any 2 points over that interval is less than 1. Then you can increase the rate of convergence of the iteration sequence with a convergence acceleration method, such as Euler Transforms, Kummer transforms, Richardson extrapolation, the e-algorithm, the Levin u-transform, Wilf-Zeilberger-Ekhad method, or the Aitken's delta-squared process (known as the Steffensen's method when applied to fixed point iterations such as this).
The last paragraph is quite difficult to carry out however, and seeing the time of your post, I am too late. Just incase you are bothered, these advanced methods have amazing rates of convergence, up to 17.83^-n with n iterations. That truly is extra ordinarily fast. In fact this just gave me an idea to calculate a sequence for pi, that will be very quick indeed. Not as quick as a professional mathematicians though, but o well.
I hope I helped.
