### Find all real solutions of \(\left(x^2-7x+11\right)^{\left(x^2-11x+30\right)}=1\)

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\[\left(x^2-7x+11\right)^{\left(x^2-11x+30\right)}=1\]

First thing I noticed was the exponent can be factorised \(x^2-11x+30=(x-5)(x-6)\)

But I don't know what to do from here

First thing I noticed was the exponent can be factorised \(x^2-11x+30=(x-5)(x-6)\)

But I don't know what to do from here

Community: Everyday Math

### 1 Answer

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Remember that any number (except 0) to the power of 0 is 1.

Which means if \(x^2-11x+30=0\) \(\left(x^2-7x+11\right)^{\left(x^2-11+30\right)}=1\)

Also remember that 1 to the power of any number is 1.

So the second case will be: \(x^2-7x+11=1\)

And there is one more case, -1 to the power of even number is 1.

So the final case will be \(x^2-7x+11=-1\), where \(x^2-11x+30\) is even

Which means if \(x^2-11x+30=0\) \(\left(x^2-7x+11\right)^{\left(x^2-11+30\right)}=1\)

Also remember that 1 to the power of any number is 1.

So the second case will be: \(x^2-7x+11=1\)

And there is one more case, -1 to the power of even number is 1.

So the final case will be \(x^2-7x+11=-1\), where \(x^2-11x+30\) is even

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