# solve exponential equations

• Nov 22nd 2008, 11:12 PM
Shyam
solve exponential equations
How would you solve the following exponential equations???

$x^x + y^y = 31$

$x^y + y^x = 17$

answer is 2 and 3. But i want to know the method to solve it.
I will be much thankful to you.
• Nov 28th 2008, 04:27 AM
fardeen_gen
I am trying to solve this problem right from the morning but I am still unsuccessful.(Worried)

Its very simple(as we have the answers) to say:
x^x + y^y = 31 = 4 + 27 = 2^2 + 3^3
x^y + y^x = 17 = 8 + 9 = 2^3 + 3^2
So comparing both sides of the equations, we see that x = 2 and y = 3

Can the MHF helpers please throw some light on what method is to be adopted to solve this type of problem?
• Dec 2nd 2008, 06:04 AM
fardeen_gen
I have scoured the internet ever since, but no successful discoveries...(Worried)
• Dec 16th 2008, 01:05 PM
Elli
• Dec 16th 2008, 01:38 PM
Isomorphism
Quote:

Originally Posted by Shyam
How would you solve the following exponential equations???

$x^x + y^y = 31$

$x^y + y^x = 17$

answer is 2 and 3. But i want to know the method to solve it.
I will be much thankful to you.

Well we can solve this equation(assuming x and y are positive integers).

Observe that x = 1 has no solution. And by symmetry y = 1 has no solution.

Thus we shall assume $x \geq 2, y \geq 2$.

But $y >3 \Rightarrow y^y > 3^3 = 27 \Rightarrow x^x+ y^y > 31$

Thus y = 2 or 3. Try both of them and get the only possible solutions $(x,y) \in \{(3,2),(2,3)\}$
• Jan 15th 2009, 06:17 PM
FreeT
Simultaneous Equation?
Considering we have 2 unknowns, and 2 equations, I would imagine that it could be solved with simultaneous equations (eventually).

Here's what I could do:

$x^x+y^y=31$
$x^x=31-y^y$
$x=\sqrt[x]{31-y^y}$

Substituting x into $x^y+y^x=17$ we get:
$\sqrt[x]{31-y^y}^{y}+y^{\sqrt[x]{31-y^y}}=17$

That's as far as I've gotten, I'll think about it more later. If anyone else has any more ideas, please post.

Hope I helped.