I don’t think this should be too hard but I’m having a harder time with it than I thought.
Let $u(w) = \frac{1}{\gamma}w^{\gamma}$
The equation I want to solve is this (where $\pi$ is a variable and not a constant).
$u(w-\pi) = \frac{1}{2}[u(w-x)+u(w+x)]$
--edit--
I’d like to solve for $\pi$
Firstly we have to insert $w-\pi$ into the utility function.
$u(w-\pi) = \frac{1}{2}[u(w-x)+u(w+x)]$
$\frac1{\gamma}(w-\pi)^{\gamma} = \frac{1}{2}[u(w-x)+u(w+x)]$
Multiplying the equation by $\gamma$.
$(w-\pi)^{\gamma} = \frac{\gamma}{2}[u(w-x)+u(w+x)]$
Taking the $\gamma$-th root on both sides of the equation.
$w-\pi = \left(\frac{\gamma}{2}[u(w-x)+u(w+x)]\right)^{\frac1\gamma}$
Multiplying the equation by $(-1)$.
$\pi-w = -\left(\frac{\gamma}{2}[u(w-x)+u(w+x)]\right)^{\frac1\gamma}$
Adding $w$.
$$\pi = w-\left(\frac{\gamma}{2}[u(w-x)+u(w+x)]\right)^{\frac1\gamma}$$
Finally you can you can replace $u(w-x)$ and $u(w+x)$ by $\frac1{\gamma}(w-x)^{\gamma}$ and $\frac1{\gamma}(w+x)^{\gamma}$, respectively.