# How to solve a quadratic raised to another quadratic?

Question:

Solve for all real $$x$$ $$(x^2 - 7x + 11)^{(x^2 - 11x + 30)} = 1$$

My thoughts:

My first thought was to take logs where the base polynomial would be the power, so $$\log_{x^2 - 7x + 11} \left( (x^2 - 7x + 11)^{(x^2 - 11x + 30)} \right) = \log_{x^2 - 7x + 11} (1)$$ Which simplifies to $$x^2 - 11x + 30 = \log_{x^2 - 7x + 11}(1)$$ This is equivalent to $$x^2 - 11x + 30 = 0$$. This can be factorised as $$(x-6)(x-5)$$. Therefore, the solutions are $$x = 5,6$$. However, it appears that I am missing one solution. How do I obtain the third solution?

• Before taking logs, one should check if the base of the log can be a valid number... Apr 28, 2023 at 3:22
• you can get another solution for when the base is $1$, namely $x^2-7x+11=1$. Apr 28, 2023 at 3:23
• also when $x^2 - 7x + 11 = -1,$ for then the exponent is an integer as well. Apr 28, 2023 at 3:26
• Apr 28, 2023 at 4:04

## 1 Answer

There are a few valid cases where a solution can exist: the base is -1, or 1, or the exponent is zero and the base is non-zero. If the base is -1, then the exponent must be even (etc.). If the base is 1, then any exponent is valid.

Then, set up three equations:

$$x^2 - 7x + 11 = 1$$

$$\to (x - 5)(x-2) = 0$$. Testing $$x = 2,5$$ in the exponent gives $$12, 0$$, and $$1^{12} = 1^0 = 1$$.

Next,

$$x^2 - 7x + 11 = -1$$

$$\to (x-4)(x-3) = 0$$. Testing $$x = 3,4$$ in the exponent gives $$6, 2$$. and $$(-1)^6 = (-1)^2 = 1$$.

Finally,

$$x^2 - 11x + 30 = 0$$

$$\to (x-5)(x-6) = 0$$. We already found $$x = 5$$, and testing $$x = 6$$ yields $$5^0 = 1$$.

Thus, the five valid solutions are $$x = 2, 3, 4, 5, 6$$.