Equation with x,y as exponential functions

In summary: Can hardly be. One equation relating two variables is a continuous curve in general. The present curve must be symmetrical about the 'diagonal' line x = y and go through that point. The curve might have more than one branch but if so no other branch cuts that line. Don't know what else they want you to say about it.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...
  • #1
cng99
44
0

Homework Statement



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


Homework Equations



That was the only equation.

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?
 
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  • #2
This is my first thread ever, by the way. Tell me if I did something wrong.
 
  • #3
cng99 said:
Solving them gives one real ordered pair of (-0.5,-0.5). But how do I know if that's the only answer?

Have you checked that that is a solution?
 
  • #4
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 realized you are supposed to describe answer geometrically, not algebraically. :wink:

Have not needed to write anything down.

Wondering whether to go back to bed or have breakfast.
 
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  • #5
epenguin said:
Have you checked that that is a solution?

(-0.5,-0.5) fits in perfectly. But the question is to find all possible real ordered pairs. How do I do that?
 
  • #6
epenguin said:
Then realized you are supposed to describe answer geometrically, not algebraically. :wink:

What do you mean?
 
  • #7
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
NascentOxygen said:
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?

I've tried that already. It would just do a differentiation. The problem can only be solved by a human.

(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
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 :redface:

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.
 
  • #11
Actually, However, that doesn't narrow down to it.
 
  • #12
epenguin said:
Then realized you are supposed to describe answer geometrically, not algebraically. :wink:

cng99 said:
What do you mean?

I mean - that I had misread the question :biggrin: (missed the ^) :blushing:
 
  • #13
cng99 said:
You get some sort of result when you click on this link, do you? As I said previously, wolframalpha seems non-functioning to me. I clicked on your link and for 10 mins watched it spinning endlessly on "Computing ..."

Given this, and in light of my inability to get it to draw even a straight-line 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 ...?"
L5zX9.gif
 
  • #14
NascentOxygen said:
Maybe this facility comes under the umbrella of "National security ...?"
L5zX9.gif


Haha! Probably.
 
  • #15
Hmmmm, no solutions then?

(-0.5,-0.5) only?
 
  • #16
cng99 said:
Hmmmm, no solutions then?

(-0.5,-0.5) only?

Can hardly be. One equation relating two variables is a continuous curve in general. The present curve must be symmetrical about the 'diagonal' line x = y and go through that point. The curve might have more than one branch but if so no other branch cuts that line. Don't know what else they want you to say about it.

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.
 
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  • #17
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
 
  • #18
cng99 said:
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

Can you show us what the family looks like, so we see what happens as you reduce this something to 1?
 
  • #19
epenguin said:
Can you show us what the family looks like, so we see what happens as you reduce this something to 1?

I can't.
 
  • #20
I've just been playing with things for a bit, and just thought I'd dumb things down for simplicity, which could also mean I dumbed things down a little too far.
If we look at the exponents only, and take the addition of them we have a function

[tex]f(x,y)=x^2+y+y^2+x[/tex]
[tex]=\left( x+\frac{1}{2} \right)^2+\left( y+\frac{1}{2} \right)^2-\frac{1}{2}[/tex]

So we clearly have a symmetry about y=x (as we already knew) and the minimum value f can take is -1/2.

At this point [itex](x,y)=(-1/2,-1/2)[/itex] which is where the minimum of f occurs, we found that

[tex]g(x,y)=16^{x^2+y}+16^{x+y^2}=1[/tex]

And so if we take some other value, say, [itex]x=\frac{-1}{2}+k[/itex], [itex]y=\frac{-1}{2}-k[/itex], [itex]k\neq0[/itex] ***
then with these values we have

[tex]x^2+y=\left(k-\frac{1}{2}\right)^2-\frac{1}{2}-k[/tex]
[tex]=\left(k-1\right)^2-\frac{5}{4}[/tex]

[tex]y^2+x=\left(k+\frac{1}{2}\right)^2-\frac{1}{2}+k[/tex]
[tex]=\left(k+1\right)^2-\frac{5}{4}[/tex]

Therefore, g becomes

[tex]g(k)=16^{(k-1)^2-5/4}+16^{(k+1)^2-5/4}[/tex]

[tex]=\frac{1}{32}\left(16^{(k-1)^2}+16^{(k+1)^2}\right)[/tex]

And for all k

[tex]16^{(k-1)^2}+16^{(k+1)^2}>16+16=32[/tex]

So we only have the minimum at (-1/2,-1/2) which means this is the only combination (x,y) that satisfies [itex]g(x,y)=1[/itex]

*** I'm uneasy about making this assumption that if we take some small linear increase of k in one variable, we should decrease the other variable with the same magnitude k.
 
  • #21
Holy Menials! Mentallic, you're a genius!
I love you!

I just thought of something new (credits go to you of course)

We know that,
Arithmetic mean >= geometric mean.

Thus,

[16^(x*x + y) + 16^(y*y + x)]/2 >= { [16^(x*x + y)]*[16^(y*y + x)] } ^1/2

Hence
g(x,y) >= 2*[16^( x*x + y*y + x + y)]^1/2

Now (x*x + y*y + x + y) >= -1/2

Thus g(x,y) >= 2 *[16^(-1/4)]

And g(x,y)>= 1

But at (x,y) = (-1/2,-1/2)
g(x,y)=1

SOLVED!I LOVE YOU!
 
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  • #22
Oh awesome! You found the answer without using any crude assumptions like I did :smile:

And I'm hardly a genius, I just gave you the kick start you needed. Using the geometric mean <= arithmetic mean was the real genius here :biggrin:
 

FAQ: Equation with x,y as exponential functions

What is an exponential function?

An exponential function is a mathematical function in which the independent variable appears in the exponent. It can be written in the form f(x) = a^x, where a is a constant known as the base. Exponential functions are commonly used to model growth or decay.

What is a system of equations with x,y as exponential functions?

A system of equations with x,y as exponential functions is a set of equations in which both the x and y variables are represented as exponential functions. This type of system can be solved by using logarithms or graphing techniques.

Can an exponential function have a negative base?

No, an exponential function cannot have a negative base. This is because raising a negative number to a power can result in a positive or negative value, making it difficult to graph and analyze. However, the base can be a fraction or decimal between 0 and 1, which would result in a decreasing exponential function.

What are the common applications of exponential functions?

Exponential functions are used in a variety of fields, such as finance, biology, and physics. Some common applications include compound interest, population growth, radioactive decay, and electronics.

How do you solve an equation with x,y as exponential functions?

To solve an equation with x,y as exponential functions, you can either use logarithms or graphing techniques. If the equation is in the form of f(x) = g(x), where both f(x) and g(x) are exponential functions, you can set them equal to each other and then use logarithms to isolate the variable. If the equation is a system of equations, you can graph both functions and find the point of intersection, which is the solution to the system.

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