**Problem of the Week**

Problem E and Solution

What are the Possibilities?

**Problem**

You may be surprised to learn that the equation \((x^2 - 5x + 5)^{x^2+4x-60} = 1\) has five solutions.

Determine all five values of \(x\) that satisfy the equation.

**Solution**

Letâ€™s consider the ways that an expression of the form \(a^b\) can be 1:

__The base, \(a\), is 1.__

In this case, the exponent can be any value and we need to solve \(x^2 - 5x + 5 = 1\).

\[\begin{aligned} x^2 - 5x + 5 &= 1\\[-1mm] x^2 - 5x + 4 &= 0\\[-1mm] (x-4)(x-1) &= 0\\[-8mm]\end{aligned}\] So \(x=4\) or \(x=1\).__The exponent, \(b\), is 0.__

In this case, the base can be any number other than 0 and we need to solve \(x^2 + 4x -60 = 0\).

\[\begin{aligned} x^2 + 4x -60 &= 0\\[-1mm] (x-6)(x+10) &= 0\\[-8mm]\end{aligned}\] So \(x=6\) or \(x=-10\).

When \(x=6\), the base is \(6^2 - 5(6) + 5 = 11 \neq 0\). That is, when \(x=6\), the base does not equal 0.

When \(x=-10\), the base is \((-10)^2 - 5(-10) + 5 = 155 \neq 0\). That is, when \(x=-10\), the base does not equal 0.

__The base, \(a\), is \(-1\) and the exponent, \(b\), is even.__

We first need to solve \(x^2 - 5x + 5 = -1\).

\[\begin{aligned} x^2 - 5x + 5 &= -1\\[-1mm] x^2 - 5x + 6 &= 0\\[-1mm] (x-2)(x-3) &= 0\\[-8mm]\end{aligned}\] So \(x=2\) or \(x=3\).

When \(x=2\), the exponent is \(2^2+4(2)-60 = -48\), which is even.

Therefore, when \(x=2\), \((x^2 - 5x + 5)^{x^2+4x-60} = 1\).

When \(x=3\), the exponent is \(3^2+4(3)-60 = -39\), which is odd.

Therefore, when \(x=3\), \((x^2 - 5x + 5)^{x^2+4x-60} = -1\). So \(x=3\) is not a solution.

Therefore, the values of \(x\) that satisfy \((x^2 - 5x + 5)^{x^2+4x-60} = 1\) are

\(x = -10\), \(x = 1\), \(x = 2\), \(x=4\) and \(x = 6\). There are five values of \(x\) which satisfy the equation.