How would you solve the following exponential equations???
$\displaystyle x^x + y^y = 31$
$\displaystyle x^y + y^x = 17$
answer is 2 and 3. But i want to know the method to solve it.
I will be much thankful to you.
I am trying to solve this problem right from the morning but I am still unsuccessful.
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?
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 $\displaystyle x \geq 2, y \geq 2$.
But $\displaystyle y >3 \Rightarrow y^y > 3^3 = 27 \Rightarrow x^x+ y^y > 31$
Thus y = 2 or 3. Try both of them and get the only possible solutions $\displaystyle (x,y) \in \{(3,2),(2,3)\}$
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:
$\displaystyle x^x+y^y=31$
$\displaystyle x^x=31-y^y$
$\displaystyle x=\sqrt[x]{31-y^y}$
Substituting x into $\displaystyle x^y+y^x=17$ we get:
$\displaystyle \sqrt[x]{31-y^y}^{y}+y^{\sqrt[x]{31-y^y}}=17$
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.