Food for thought

## Solution

1. Find all real solutions of the equation ${\large (x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1}.$

There are several possibilities here.

Case 1. For any number $a$ we have $a^0=1$, so if we solve $x^2-11x+30=0$ then this will give us some solutions.

Case 2. For any number $b$ we have $1^b=1$, so if we solve $x^2-7x+11=1$ we will find more solutions.

Case 3. We also have that $(-1)^{2b}=1$ for any number $b$. So we can solve $x^2-7x+11=-1$, and then see whether the numbers we get lead to $x^2-11x+30$ being an even number.

These are the only options, so by dealing with each case in turn we’ll find all the solutions.

For Case 1, we get $x^2-11x+30=(x-6)(x-5)=0,$ so $x=5$ and $x=6$ are solutions.

For Case 2, we solve $x^2-7x+10=(x-2)(x-5)=0,$ so we have an additional solution, $x=2$.

Finally, for Case 3 we have that $x^2-7x+12=(x-3)(x-4)=0,$ so we must check: \begin{align*} 3^2-11\cdot 3+30=6 \quad \checkmark\\ 4^2-11\cdot 4 +30=2 \quad \checkmark \end{align*}

are both even, so $x=3$ and $x=4$ are also solutions.

So the real numbers satisfying the equation are $x=2,3,4,5, 6.$

1. Find all real solutions of the equation ${\large (2-x^2)^{(x^2 - 3\sqrt{2}x + 4)} = 1}.$

There are again three cases to consider:

Case 1. $x^2-3\sqrt{2}x+4=0$,

Case 2. $2-x^2=1$,

Case 3. $2-x^2=-1$ and $x^2-3\sqrt{2}x+4$ is even.

In Case 1, $x^2-3\sqrt{2}x+4=(x-\sqrt{2})(x-2\sqrt{2})=0,$ and so we have that $x=\sqrt{2}$ and $x=2\sqrt{2}$ are solutions. However, when $x=\sqrt{2}$, we end up with $0^0$.

The quantity $0^0$ is a bit problematic. For some mathematicians there are good reasons to define $0^0$ to be $1$, whereas other mathematicians prefer to say that $0^0$ is undefined.

So we may choose to discard $x=\sqrt{2}$ as a solution.

In Case 2, we have that $x^2-1=(x+1)(x-1)=0,$ so $x=1$ and $x=-1$ are also solutions.

Solving Case 3, we get $x^2-3=(x+\sqrt{3})(x-\sqrt{3})=0,$ and so $x=\pm\sqrt{3}$ are potential solutions. However, we have that $\sqrt{3}^2\pm3\sqrt{2}\sqrt{3}+4=7\pm3\sqrt{6}$ is not divisible by two (it isn’t even an integer!) so $x=\pm\sqrt{3}$ are not solutions.

So the values of $x$ satisfying the equation are $x=-1, 1, 2\sqrt{2}.$