SOLUTION: How do I solve (x^2-5x+5)^(x^2-9x+20)=1?

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Question 250958: How do I solve (x^2-5x+5)^(x^2-9x+20)=1?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This one is a bit tricky. We have to think out how we can take a number and reise it to a power and get a result of 1. There are three possibilities I can think of:
  • 1 to any power is 1
  • Any number (excpet zero) to the zero power is 1
  • -1 to any even power is 1

So we will look for solutions from all three possibilities.
1 to any power is 1:
Solve x%5E2+-5x+%2B+5+=+1
Subtract 1 from each side:
x%5E2+-5x+%2B+4+=+0
Factor:
x+-4%29%28x-1%29=+0
Use the Zero Product Property:
x+-4+=+0 or x+-+1+=+0
Solve:
x+=+4 or x+=+1

Any number but zero to the zero power is 1
For this we will find the x's that make x%5E2+-9x+%2B20+=+0 and then check to make sure they do not make x%5E2-5x%2B5 zero, too. Solving x%5E2+-9x+%2B20+=+0...
Factor:
%28x-4%29%28x-5%29+=+0
Use the Zero Product Property:
x-4+=+0 or x-5+=+0
Solve:
x+=+4 or x+=+5
Now we check to see if either of these make x%5E2+-+5x+%2B5 zero:
for x = 4:
%284%29%5E2+-+5%284%29+%2B+5
16+-+5%284%29+%2B+5
16+-+20+%2B+5
1 This is not 0 so x = 4 is OK.
for x = 5:
%285%29%5E2+-+5%285%29+%2B+5
25+-+5%285%29+%2B+5
25+-+25+%2B+5
5 This is not 0 so x = 5 is OK.

-1 to an even power is 1
Solve x%5E2+-5x+%2B+5+=+-1 and then see if x%5E2+-9x+%2B+20 is even for those x's.
Solve x%5E2+-5x+%2B+5+=+-1:
Add 1 to each side:
x%5E2+-5x+%2B+6+=+0
Factor:
%28x-3%29%28x-2%29+=+0
Use the Zero Product Property:
x=3+=+0 or x-2+=+0
Solve:
x = 3 or x = 2
Now we will see if either of these make the exponent, x%5E2+-9x+%2B+20, even:
for x = 3:
%283%29%5E2+-9%283%29+%2B+20
9+-9%283%29+%2B+20
9+-27+%2B+20
2 This is even so x = 3 is a solution.
Now we will see if either of these make the exponent, x%5E2+-9x+%2B+20, even:
for x = 2:
%282%29%5E2+-9%282%29+%2B+20
4+-9%282%29+%2B+20
4+-18+%2B+20
2 This is also even so x = 2 is also a solution.

So the solutions are: x = 4, x = 1, x = 5, x = 3 and x = 2.