Question 422017: (x-2)^(x^2-x-12)=1
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! When I first looked at your problem I thought it would be difficult. But after a few moments I realized that it is a lot easier than it looks.
This problem is easy because there are only three ways to raise a number to a power and get a result of 1:- Raising a 1 to any power always results in a 1
- Raising a -1 to any even power results in a 1
- Raising any number (except zero) to the zero power results in a 1
Let's look at all three possibilities:
1) Raising 1 to a power.
The base in your expression is x-2. If x = 3 then the base would be a 1. And since 1 to any power is a 1, it doesn't matter what the exponent works out to be. So x = 3 is a solution to your equation.
2) Raising -1 to an even power.
The base is still x-2. If x = 1 then the base would be -1. If x = 1, would the exponent,  , be even? Let's see:
Yes, if x = 1 the base is -1 and exponent is an even number. So x = 1 is another solution to your equation.
3) Raising a non-zero number to the zero power.
First let's find the x's that make the exponent, zero:
As a quadratic equation we will either factor this or use the Quadratic Formula. This factors fairly easily:
(x-4)(x+3) = 0
From the Zero Product Property we know that one of the factors must be zero. So:
x-4 = 0 or x+3 = 0
Solving these we find:
x = 4 or x = -3
Now we just have to make sure the base is not zero when x is one of these numbers. When x = 4, the base, x-2, is a 2. And when x = -3, the base is -5. So when x = 4 or x = -3, the exponent is zero and the base is not. So x = 4 or x = -3 are also solutions to this equation.
So there are four solutions to your equation:
x = 3 or x = 1 or x = 4 or x = -3
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