LollyLau
  • LollyLau
Series of trial-and-error questions: Find all the unknowns for: 1. x^y=y^x 2. x^y=y^x+1 where all the unknowns are integers.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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LollyLau
  • LollyLau
?????????!!!!!!!!
anonymous
  • anonymous
i guess \[y=x\] might work for first one
LollyLau
  • LollyLau
could be...

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More answers

LollyLau
  • LollyLau
let's say they aren't, some actual answers please...
anonymous
  • anonymous
what is wrong with \[3^3=3^3\]?
anonymous
  • anonymous
2 and 4 works for the first one.
LollyLau
  • LollyLau
Nothing! Pratu got it right :) (Well you too...)
Mr.Math
  • Mr.Math
Sorry, I thought it was a system of two equations.
LollyLau
  • LollyLau
oh.
anonymous
  • anonymous
ooooooooooooooh. i was wondering...
Mr.Math
  • Mr.Math
I agree with satellite on the first one, the solution of the first one is all integers (x,y) such that x=y. For the second one, we have (x,y)=(2,1) is a solution.
Mr.Math
  • Mr.Math
(3,2) is another solution.
anonymous
  • anonymous
pratu had \[2^4=4^2\] so that works as well. my guess is there are no others, but that is a guess
LollyLau
  • LollyLau
There are no others.
LollyLau
  • LollyLau
no proof tho. just know that for SURE :)
LollyLau
  • LollyLau
Mr. Math, you're missing one more answer. (which i know)
LollyLau
  • LollyLau
(solution)
LollyLau
  • LollyLau
ans. in half a min.
Mr.Math
  • Mr.Math
(1,0).
LollyLau
  • LollyLau
YAY!
Mr.Math
  • Mr.Math
Actually any solution in the form (x,0), x is a positive integer.
LollyLau
  • LollyLau
Can we do this for +2, +3, +4 etc.
Mr.Math
  • Mr.Math
Yes.

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