WMC 2018 Senior Problem 37

First, rewrite the equation using the laws of exponents: $x^{\frac{10+x}{x}} = x^{x-2}$ We are left with three cases:

• $x=0$ : this is impossible, as it would make the fraction $\frac{10+x}{x}$ undefined.
• $x=\pm 1$ : Both of these are solutions: $\begin{aligned} 1^{11} &= 1^{-1} \\ (-1)^{-9} &= (-1)^{-3} \end{aligned}$
• $\frac{10+x}{x} = x-2$ : this case results from equating the exponents on both sides of the equation. Multiplying by $x$ yields $\begin{aligned} 10+x &= x^2 - 2x \\ 0 &= x^2 -3x - 10 \\ 0 &= (x-5)(x+2) \end{aligned}$ Thus $x=-2$ and $x=5$ are solutions as well (and, indeed, plugging them in yields $(-2)^{-4} = (-2)^{-4}$ and $5^3 = 5^3$ ).

The sum of the solutions is $1 - 1 + 5 - 2 = \boxed{3}$ .