- 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 \[x=2,3,4,5, 6.\]
- 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 \[x=-1, 1, 2\sqrt{2}.\]
