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?
You missed some cases.
The cases are: $\text{number}^0,1^\text{number},{-1}^{\text{even}}$
You have only considered the first case.
For first case, $$x^2+4x-60=0$$ the roots are $6,-10$.
For the second case, $$x^2-5x+5=1$$ the roots are $1,4$.
For the third case, $$x^2-5x+5=-1$$ $x$ can be $2$ or $3$. If you substitute the value $2$ on the equation $x^2+4x-60$, you get $-48$ which satisfies the condition since $-1^{-48}=1$. If you substitute $x=3$, you get $-39$ which does not satisfy the given condition since $-1^{-39}\neq 1$. So only $x=2$ satisfies the condition.
The real values of $x$ are therefore: $1,2,4,6,-10$. The sum of real values of $x$ is therefore $3$.
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.
