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

Let’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.