<s>[INST] Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)

Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.

My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ and hence the equation becomes one. So, the sum of the real values of $x$ satisfying the equation is $6+(-10)=-4$. But answer is $+3$.
Where have I gone wrong?
 [/INST] First of all, the equation in your link is : $(x^2-5x+5)^{x^2+4x-60}=1$.
Simply use $\ln$ in both sides :
$ \ln(x^2-5x+5)^{x^2+4x-60} = \ln1 \Leftrightarrow (x^2 + 4x - 60)\ln (x^2 - 5x + 5) = 0$.
That means that : $x^2 + 4x - 60 = 0$ or $(x^2 - 5x + 5) = 1$.
Of course, the functions inside of $\ln$ must be bigger than zero, do not forget to add that in your final answer after you've done these.
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