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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?

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1 Answer 1

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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 $x = 2, 3, 4, 5, 6$.

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