SuicideFuel Problem

[trying to ascend]

16 to the power of (x² + y) + 16 to the power of (y² + x) = 1.

What are all the possible values of x and y?

 

[CroatianManlet]

No idea let me know when you find the answer

 

[NoLooksNoLife]

Krempee

 

[Fancy Alcoholic]

Believing in maths is trucel trait. Chad doesn't have to solve X to get a gorillon of pussy.

 

[BornToLose]

I'm sick of your problems. Get a real problem already.

 

[racoon4]

(16^x²+y) + (16^y²+x) = 1. What are the values of x and y?

Answer (1 of 5): 16^(x²+y) + 16^(y²+x) = 1 …….(1) Equation (1) is only and only possible when 16^(x²+y) = 1/2 and 16^(y²+x) = 1/2 Then 1/2 + 1/2 = 1 Take 1st part… 16^(x²+y) = 1/2 [{2}⁴]^(x²+y) = 2^(-1) 2^{4(x²+y)} = 2^(-1) Comparing power Therefore, 4(x²+y) = -1 x²+y = -1/4 x² + y + 1/...

www.quora.com

 

[TheCatMan]

4

 

[NoBitches]

16 to the power of (x² + y) + 16 to the power of (y² + x) = 1.

What are all the possible values of x and y?

So from what I understand 16^[(x^2+y)+16]^(y^2+x)=1
16^0=1
Which means that
((x^2+y)+16)^(y^2+x)=0
Which just means that ((x^2+y)+16)=0 as it cant end up with a value of zero otherwise
Which just gives us y=-16-x^2
So a simple value that works is x=0 and y=-16 although there are infinite solutions if I am reading this correctly

 

[trying to ascend]

So from what I understand 16^[(x^2+y)+16]^(y^2+x)=1
16^0=1
Which means that
((x^2+y)+16)^(y^2+x)=0
Which just means that ((x^2+y)+16)=0 as it cant end up with a value of zero otherwise
Which just gives us y=-16-x^2
So a simple value that works is x=0 and y=-16 although there are infinite solutions if I am reading this correctly

Wrong