Going gets tough, tough resort to numerical methods. The following is Mathematica code to calculate the value of n for the supplied values of R and L. First out is the expression for the sum, second out is the numerically calculated value of n which makes the sum equal to zero, and the third out is a back-substitution of the the root for a check.
Code:
In[234]:=
R = 3;
L = 5;
formula1 = Sum[Binomial[R, r]*(-1)^(R - r)*
(r/L)^n*Log[r/L], {r, 1, R}]
root = n /. FindRoot[formula1 == 0,
{n, 1.2}]
formula1 /. n -> root
Out[236]= -(3/5)^n Log[5/3] + 3 (2/5)^n Log[5/2] - 3 5^-n Log[5]
Out[237]= 1.38015
Out[238]= -5.55112*10^-17