3. Problem: Find the sum of the smallest and largest possible values for $x$ which satisfy the following equation.
$$
9^{x+1}+2187=3^{6 x-x^{2}}
$$
Solution: (Viet Hoang)
First we prove that the given equation has at least one root. To do this we consider the following function.
$$
f(x)=9^{x+1}+2187-3^{6 x-x^{2}}
$$

Note that this function is continuous. Note also that $f(0)=9+2197-1>0$ and $f(3)=$ $9^{4}+2187-3^{9}<0$. Therefore by the intermediate value theorem there exists a real number $0<z<3$ such that $f(z)=0$. Therefore there is at least one value $x$ which statisfies the equation.

Next, we can rewrite the equation in the following form
$$
3^{x}+3^{5-x}=3^{x(5-x)-2}
$$

Let $r$ and $s$ be any two numbers such that $r+s=5$. Notice that if $r$ is a root of 1 then $s$ must also be a root (and vice versa). This is because
$$
3^{r}+3^{5-r}=3^{5-s}+3^{s} \quad \text { and } \quad 3^{r(5-r)-2}=3^{(5-s) s-2} .
$$

Now we claim that if $x_{0}$ is the smallest root of 1 , then $\left(5-x_{0}\right)$ must be the largest root. Indeed if $x_{1}$ were a root of 1 greater than $\left(5-x_{0}\right)$, then $\left(5-x_{1}\right)$ would also have to be a root, but $\left(5-x_{1}\right)<\left(5-\left(5-x_{0}\right)\right)=x_{0}$ and this would contradict the fact that $x_{0}$ is the smallest root. Therefore, the sum of the largest and the smallest roots of 1 is $x_{0}+\left(5-x_{0}\right)=5$.