Question

I need to find 5 solutions to this problem...

Answer 1

\[(x^2-5x+5)^{(x^2-9x+20)}=1 \] I attempted it so don't say "please try to attempt it" or something.

Answer 2

I also know that any number with the power of zero is one.. but eh.

Answer 3

Are you SURE there are 5?

Exponent is Zero but Base is NOT zero.
\(x^{2} - 9x + 20 = 0\)
\((x-5)(x-4) = 0\)
\(x = 5\;or\;x = 4\) -- There's two. unless \(5^{2} - 5(5) + 5 = 0\;or\;4^{2} - 5(4) + 5 = 0\). Neither is zero, so we're good.

Base is 1 but exponent is Finite
\(x^{2} - 5x + 5 = 1\)
\(x^{2} - 5x + 4 = 0\)
\((x-4)(x-1)=0\)
\(x = 4\;or\;x = 1\) -- Too bad. A repeat. We're up to 3!

Answer 4

Yeah, there's 5 solutions unfortunately :(

Answer 5

what happens if you try 2 ?

Answer 6

if you get a base of \(-1\) and the exponent of something even, then it will work i think

Answer 7

\[x^2-5x+5=-1\]
\[x^2-5x+6=0\]
\[(x-2)(x-3)=0\] so try \(x=2\) and see if the exponent is an even integer

Answer 8

oh yeah it has be be even, because \(2\) is even
so \(2\) will work

Answer 9

Both are strictly positive and greater than unity for x < 1 - There cannot be solutions out there.

Both are strictly positive and greater than unity for x > \((1/2)(9+\sqrt{5}\) - There cannot be solutions out there.

Anything interesting has to be in \([1,5.618]\)

Answer 10

\(x=2\) works for sure
base is \(-1\) an exponent is even

Answer 11

maybe you will get lucky and 3 will work also
base is \(-1\)

Answer 12

yeah it works too

Answer 13

5, 4, 1, 2 and 3? Those are the five solutions?

Answer 14

Very nice. It's not yet clear to me why I can't get my software to graph them. I pulled all the usual tricks for negative values and logarithms. It will take a little thought.

Answer 15

Okay, I'm going to check them all right now.

Answer 16

Wait.. I used 2 and got -1.. will that work?

Answer 17

Thanks you guys, I wish that I could give both of you medals. :(