find

vijay1729 · **Joined:** Wed, 30 Nov 2011 09:40:13 UTC **Posts:** 5

Find the all real x,y such that 16^(x^2+y)+ 16^(y^2+x) =1

outermeasure · **Posted:** Wed, 30 Nov 2011 11:06:38 UTC

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.

Soroban · **Posted:** Wed, 30 Nov 2011 16:48:30 UTC

Hello, vijay1729!

Quote:

$\text{Find the all real }x,y\text{ such that: }\:16^{x^2+y}+ 16^{y^2+x} \:=\:1$

$\text{Due to the symmetry of the }x\text{'s and }y\text{'s (they are interchangeable),}$
. . $\text{we can conclude that }\,x = y.$

$\text{and we have: }\:2\cdot16^{x^2+x} \:=\:1\quad\Rightarrow\quad 2\cdot (2^4)^{x^2+x} \:=\:1$

. . . . . . . . . . . $2\cdot2^{4x^2+4x} \:=\:1 \quad\Rightarrow\quad 2^{4x^2 + 4x+1} \:=\:2^0$

$\text{Hence: }\:4x^2 + 4x + 1 \:=\:0 \quad\Rightarrow\quad (2x+1)^2 \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}\frac{1}{2}$

$\text{Therefore: }\:(x,y) \;=\;\left(\text{-}\frac{1}{2},\:\text{-}\frac{1}{2}\right)$

outermeasure · **Posted:** Wed, 30 Nov 2011 16:56:12 UTC

Soroban wrote:

Hello, vijay1729!

Quote:

$\text{Find the all real }x,y\text{ such that: }\:16^{x^2+y}+ 16^{y^2+x} \:=\:1$

$\text{Due to the symmetry of the }x\text{'s and }y\text{'s (they are interchangeable),}$
. . $\text{we can conclude that }\,x = y.$

$\text{and we have: }\:2\cdot16^{x^2+x} \:=\:1\quad\Rightarrow\quad 2\cdot (2^4)^{x^2+x} \:=\:1$

. . . . . . . . . . . $2\cdot2^{4x^2+4x} \:=\:1 \quad\Rightarrow\quad 2^{4x^2 + 4x+1} \:=\:2^0$

$\text{Hence: }\:4x^2 + 4x + 1 \:=\:0 \quad\Rightarrow\quad (2x+1)^2 \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}\frac{1}{2}$

$\text{Therefore: }\:(x,y) \;=\;\left(\text{-}\frac{1}{2},\:\text{-}\frac{1}{2}\right)$

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).

vijay1729 · **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.

outermeasure · **Posted:** Thu, 1 Dec 2011 16:10:13 UTC

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 f(x)=2^x