Equation with x,y as exponential functionsby cng99 Tags: exponential equation, olympiad, polynomial exponent, two variable, x y 

#1
Mar812, 11:52 PM

P: 44

1. The problem statement, all variables and given/known data
This question was asked in the Indian National Maths Olympiad. The question is: Find all the possible real ordered pairs of (x,y) for equation 16^[(x^2) + y] + 16^[x + (y^2)]=1 2. Relevant equations That was the only equation. 3. The attempt at a solution Using 0.5 + 0.5 =1, (x^2) + y = 0.25 and (y^2) + x= 0.25 (Because [16^(0.25)]=0.5 , I'm using decimals here instead of fractions) Solving them gives one real ordered pair of (0.5,0.5). But how do I know if that's the only answer? 



#2
Mar812, 11:54 PM

P: 44

This is my first thread ever, by the way. Tell me if I did something wrong.




#4
Mar1012, 01:20 AM

HW Helper
P: 1,932

Equation with x,y as exponential functions
Was up till 3 a.m. here last night for a family arrival, and wrote last post just before going to bed. Had thought to write 'is it plausible that the answer is one or a few pairs of numbers like that?'.
Woke about 7, and in a bit saw ± what you are supposed to do. It starts looking nice but at the end the algebra got looking ugly. Then realised you are supposed to describe answer geometrically, not algebraically. Have not needed to write anything down. Wondering whether to go back to bed or have breakfast. 



#5
Mar1112, 12:01 AM

P: 44





#6
Mar1112, 12:03 AM

P: 44





#7
Mar1112, 12:19 AM

HW Helper
P: 4,715

I tried over two days to see what wolframapha would make of this, but it's not functioning. Anyone know whether the free version has been decommissioned?




#8
Mar1112, 12:57 AM

P: 44

(It might be possible to plot a graph of 16^((x^2) + y) + 16^((y^2) + x) = z, and then intersecting the surface with z=1. But I couldn't find anything that would plot something like that. They all accept y as a function of x. This doesn't really get us anywhere. ) Is there anyway to find the minimum or maximum value of 16^((x^2) + y) + 16^((y^2) + x)? That might help. Possible ways to do so might be using the Arithmetic Mean >= Geometric mean >= Harmonic mean property. Or maybe by just differentiating it and putting dy/dx=0. But I couldn't find anything there. Do tell me if anyone finds anything. 



#9
Mar1112, 01:15 AM

HW Helper
P: 3,436

I'm having quite a lot of trouble with this problem, it's really got me stumped. But then again, all the Olympiad questions seem to give me a hard time
Since we have an exponential of the form [itex]a^b+a^c=1[/itex] and we know that [itex]a^b>0, a^c>0[/itex] then what we must have is that [itex]b<0, c<0[/itex] So [tex]x^2+y<0[/tex] [tex]x+y^2<0[/tex] And this gives us the inequality [tex]1<x<0[/tex] [tex]\sqrt{x}<y<x^2[/tex] But from here I can't think of a way to further narrow down our possible solution set. 



#10
Mar1112, 03:20 AM

P: 44

Hey, Mentallic. Nice try...
This did help a bit... http://www.wolframalpha.com/input/?i...B+1%3C+y+%3C0 



#11
Mar1112, 03:27 AM

P: 44

Actually, However, that doesn't narrow down to it.




#13
Mar1112, 07:17 AM

HW Helper
P: 4,715

Given this, and in light of my inability to get it to draw even a straightline graph, I think the owners must have made it regionally selective, so it's ignoring certain IP ranges. When I go to http://www.wolframalpha.com I am presented with a totally blank page. Anyone else? Maybe this facility comes under the umbrella of "National security ..........?" 



#14
Mar1112, 10:47 AM

P: 44

Haha! Probably. 



#15
Mar1412, 04:26 AM

P: 44

Hmmmm, no solutions then?
(0.5,0.5) only? 



#16
Mar1512, 05:47 AM

HW Helper
P: 1,932

You might therefore try rotating the thing counter clockwise by 45° , then it is symmetrical around the new X axis f(X) = f(X) and see if that inspires anything. 



#17
Mar1512, 07:42 AM

P: 44

Well, if it's a curve, why wouldn't anyone plot it?
Wolfram Alpha plots a curve for 16^[(x^2) + y] + 16^[x + (y^2)]= anything more than 1. Maybe whatever the curve is, it gets reduced to a point when 16^[(x^2) + y] + 16^[x + (y^2)]= 1 


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