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 Post subject: findPosted: Wed, 30 Nov 2011 09:43:43 UTC

Joined: Wed, 30 Nov 2011 09:40:13 UTC
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Find the all real x,y such that 16^(x^2+y)+ 16^(y^2+x) =1

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 Post subject: Re: findPosted: Wed, 30 Nov 2011 11:06:38 UTC
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vijay1729 wrote:
Find the all real x,y such that 16^(x^2+y)+ 16^(y^2+x) =1

(x,y)=(-1/2,-1/2) is the only solution.

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 Post subject: Re: findPosted: Wed, 30 Nov 2011 16:48:30 UTC
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Hello, vijay1729!

Quote:

. .

. . . . . . . . . . .

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 Post subject: Re: findPosted: Wed, 30 Nov 2011 16:56:12 UTC
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Soroban wrote:
Hello, vijay1729!

Quote:

. .

. . . . . . . . . . .

No! You can't conclude x=y just because the LHS is symmetric in x and y. For example, the same LHS but with RHS=32 obviously has solution that breaks the symmetry (x,y)=(0,1) or (1,0).

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 Post subject: Re: findPosted: Thu, 1 Dec 2011 06:50:32 UTC

Joined: Wed, 30 Nov 2011 09:40:13 UTC
Posts: 5
yes If the equation is symmetric in x,y then one of the solution is x=y some times this cant be solution also, so we cannot conclude that x=y with out proper argument, If we prove that there is no solution in x>y or x<y i think we can conclude x=y=-1/2 is the only solution.

any body solved please post here.

i want know how to solve this problem.

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 Post subject: Re: findPosted: Thu, 1 Dec 2011 16:10:13 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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vijay1729 wrote:
yes If the equation is symmetric in x,y then one of the solution is x=y some times this cant be solution also, so we cannot conclude that x=y with out proper argument, If we prove that there is no solution in x>y or x<y i think we can conclude x=y=-1/2 is the only solution.

any body solved please post here.

i want know how to solve this problem.

I don't want to deny you the pleasure of solving it yourself, so I'll just erect pointers.

Let , . We want to solve .

Since f is nonnegative convex and turns addition to multiplication, with unless , we have , with equality if and only if .

Now what is ?

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