How to solve the transcendental equation $\frac{\zeta'(\alpha)}{\zeta(\alpha)}=-\frac{1}{n}\sum_{i=1}^n\ln x_i$ How can I solve the transcendental equation: 
$$\frac{\zeta'(\alpha)}{\zeta(\alpha)}=-\frac{1}{n}\sum_{i=1}^n\ln x_i,$$
where $\displaystyle\zeta(\alpha)=\sum_{m=1}^\infty m^{-\alpha}\;\;\;?$ The values $x_i \in \{1, 2, 3, ..., 63, 64\}$ are given so the solution we are looking for, is $\alpha^*$ that satisfies this equation. This equation is from Clauset's et al paper: "Power-law distributions in empirical data". In other words the problem is: Solve $\alpha$ that satisfies: 
$$\frac{\zeta'(\alpha)}{\zeta(\alpha)}=C,$$
where $C$ is constant. 
What methods / algorithms should I use? I've been told this should be solved numerically, but I don't explicitly know what method I should use. This equation arises from solving the parameters of discrete power-law distribution. 
 A: There are no analytical solution to your equation so you must use some numerical scheme. There are many methods to find roots of a function. The simplest one (and usually good enough for most purposes) are

*

*Bisection

*Newton's method / Secant method
I explained the bisection method and Newton's method in this answer. If you are to implement a root finding method yourself then I would suggest you start with bisection: it's the easiest one to implement and there is little that can go wrong (less chance for making mistakes).
If you are allowed to use mathematical software there are many built-in methods (implementing methods like the ones mentioned above). Always use built-in methods if you can. Below are some examples for how you can find a root of a function in some of the most common packages Matlab, Mathematica and Maple. The example is finding the root of $\sin(x)$ close to $x=x_0=3$ (which should give $x=\pi$).
Mathematica's FindRoot:
f     = Sin[x];
x0    = 3;
FindRoot[f[x] == 0, {x, x0}]

MatLab's fzero:
f     = @sin;
x0    = 3;
fzero(f, x0)

Maple's fsolve:
f    := sin(x);
fsolve(f = 0, x, 2..4);


For your spesific question: The function $-\frac{\zeta'(x)}{\zeta(x)}$ is positive and monotonely decreasing on $(1,\infty)$ and asymptotes to $0$ so the function $F(x) = -\frac{\zeta'(x)}{\zeta(x)}-C$ has only one root for any $C > 0$ (see plot below). Any of the methods mentioned above should work to find this root. Using the asymptotics of the $\zeta$-function we can derive approximate solutions for the roots in the limit $C\gg 1$ and $C\ll 1$.
Since $\zeta(x) \approx \frac{1}{x-1}$ close to $x=1$ we get that the root is approximately
$$x \approx \frac{1}{C}~~~~~~\text{for}~~~~~~C\gg 1$$
For large $x$ we have $\zeta(x) \approx 1 + \frac{1}{2^x}$ which gives us that the root is approximately
$$x \approx \frac{\log\left(\frac{1}{C}\right)}{\log(2)}~~~~~~\text{for}~~~~~~C\ll 1$$

