Please help

**dapore** · December 16th 2009, 03:48 AM

Find the solution with the reasoning

x^y=8 , y^x=9

**Raoh** · December 16th 2009, 03:54 AM

Originally Posted by dapore

Find the solution with the reasoning

x^y=8 , y^x=9

hi

$x^y=2^3=8,y^x=3^2=9$ .
therefore,
$x=2$ and $y=3$ .

**dapore** · December 16th 2009, 03:59 AM

Thank you very much, but how to prove that this solution is the only solution

**Bacterius** · December 16th 2009, 04:41 AM

Can you find another (integer I believe) combination of $x$ and $y$ that satisfy $x^y = 8$ ?

**Raoh** · December 16th 2009, 05:14 AM

Originally Posted by Bacterius

Can you find another (integer I believe) combination of $x$ and $y$ that satisfy $x^y = 8$ ?

is there anyway we can prove it

,or is it just obvious.

**Raoh** · December 16th 2009, 05:18 AM

i was just wondering how much solutions this equation has,

$y^x-x^y=1$

**dapore** · December 16th 2009, 02:47 PM

Raoh
my full gratitude and appreciation

**dapore** · December 18th 2009, 06:15 AM

This my way of proof

1 - can not be any of the X, Y number is negative because in this case it would output the left end in a fraction equations absolute value smaller than 1
2 - can not be any of the X, Y equal to zero because in this case would be the left end result = either 1 or zero or undefined
3 - can not be any of the X, Y equal to 1 because it has achieved one of the equations can not be achieved while the other
And therefore there is no option for X, Y only be a Moajabin greater than 1
4 - can not be greater than Q Ooisawi 3 because when you is no more achieved the first equation and therefore the only value for Q is 2
5 - can not be greater than p. 4 Ooisawi because it is no more realized when the second equation and therefore the only value of r is 3

What do you think

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