Closed form solution for constant exponent in sum I am trying to solve for $\alpha$ in the following equation:
$$ 0.80 = \frac{1}{3} \left( X_1^\alpha + X_2^\alpha + X_3^\alpha \right)$$
Right now I just use Excel and solver to find a numerical solution to problems like these.  Seems like there should be a way to obtain a closed form solution.
This is an exponential curving example for grades that sets the average post-curved grade to 80%. Let me make it a bit more general:
$$ 0.80 = \frac{1}{n} \sum_i^n X_i^\alpha$$ where $n$ is the number of students that took the exam, $0 < X_i < 1$ is an exam grade for student $I$, and $0 < \alpha \le 1$ is a scaling factor.  Everything is known except $\alpha$.
This type of curving "gives to the needy and not the greedy" in that lower scoring students receive more of a bump yet the rank ordering is unchanged.

Considering that you look for the zero of function
$$f(\alpha)=\sum_{i=1}^n X_i^\alpha - nk$$ without loss of generality, suppose that the $X_i$ are such that $X_1\leq X_2\leq \cdots \leq X_n$.
So the solution is bounded
$$ \alpha_{min}=\frac{\log (k)}{\log (X_1)}\leq \alpha \leq  \frac{\log (k)}{\log (X_n)} =\alpha_{max}$$ with $f( \alpha_{min})>0$ and  $f( \alpha_{max})<0$.
Since $$f''(\alpha)=\sum_{i=1}^n X_i^\alpha \log ^2(X_i)$$ is positive, start Newton method with $\alpha_0=\alpha_{min}$ and, by Darboux theorem, you will reach the solution without any overshoot of the solution.
Let us try with $n=6$, $k=0.8$ and $\{0.123,0.234,0.235,0.456,0.567,0.678\}$ for the $X_i$. You should get the following iterates
$$\left(
\begin{array}{cc}
n & \alpha_n \\
 0 & 0.106483 \\
 1 & 0.198764 \\
 2 & 0.205234 \\
 3 & 0.205263
\end{array}
\right)$$
For the fun of it, let us use $n=1000$ the $X_i$'s being normally distributed between $0.2$ and $0.8$ $(\mu=\frac 23,\sigma=\frac 13)$. This would give the following iterates
$$\left(
\begin{array}{cc}
n & \alpha_n \\
 0 & 0.165337 \\
 1 & 0.348781 \\
 2 & 0.362371 \\
 3 & 0.362439 
\end{array}
\right)$$
