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February 22nd, 2017, 06:44 AM   #1
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How could I solve this system of equations???

Solve:
$\displaystyle \left\{\begin{matrix}
&{\sqrt{2^{2^{x}}}}= 2\left(y+1\right) \\
& {\sqrt{2^{2^{y}}}}= 2\left(x+1\right)
\end{matrix}\right\}$.

Last edited by skipjack; February 22nd, 2017 at 08:17 AM.
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February 22nd, 2017, 10:13 AM   #2
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Initial reaction:
Obvious that x = y.
So only one of the equation needs to be solved.

How wrong am I?

Anyhooooooo: x=y=3.

If y=3, then:
x = log[4(y^2 + 2y +1)] / {2[log(2)]} = 3
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February 22nd, 2017, 10:49 AM   #3
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Quote:
Originally Posted by Denis View Post
Initial reaction:
Obvious that x = y.
So only one of the equation needs to be solved.

How wrong am I?

Anyhooooooo: x=y=3.

If y=3, then:
x = log[4(y^2 + 2y +1)] / {2[log(2)]} = 3
I quit...
Umm

$x = 3 \implies \sqrt{2^{2^3}} = \sqrt{2^8} = 2^4 = 16 \implies$

$2(y + 1) = 16 \implies y + 1 = 8 \implies y = 7 \ne 3.$

$y = 7 \implies \sqrt{2^{2^7}} = \sqrt{2^{128}} = 2^{64}.$

$2(x + 1) = 2^{64} \implies x = 2^{63} - 1 \ne 3.$
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February 22nd, 2017, 11:00 AM   #4
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If the right-hand sides of the equations were $2^{y+1}$ and $2^{x+1}$, $x=y=3$ would work, but wouldn't be the only solution.
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February 22nd, 2017, 11:02 AM   #5
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Bad eyes...I read OP as 2^(2*x)

Yer peepers better than mine!
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February 22nd, 2017, 11:06 AM   #6
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Quote:
Originally Posted by Denis View Post
Initial reaction:
Obvious that x = y.
So only one of the equation needs to be solved.

How wrong am I?

Anyhooooooo: x=y=3.
nuuuu

using your keen insight above we have to solve

$\sqrt{2^{2^x}} = 2(x+1)$

Mathematica says no closed form solutions for you!

A quick graph and numerical solution later reveals

$x=y \approx 2.49$

$x=y \approx -0.343$
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February 22nd, 2017, 11:23 AM   #7
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OK; so, using 1st equation:

u = log(2), v = log(4y^2+8y+4)

-x = log(u/v) / u

Oui?
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February 22nd, 2017, 12:01 PM   #8
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Quote:
Originally Posted by Denis View Post
OK; so, using 1st equation:

u = log(2), v = log(4y^2+8y+4)

-x = log(u/v) / u

Oui?
happy hour start early today?

you abandoned your key insight that $x=y$ !
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February 22nd, 2017, 12:05 PM   #9
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I find it can be rearranged into the following

$\large 2=(x+1)^{\frac{1}{2^{x-1}-1}}$

but it still has to be numerically solved.
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February 22nd, 2017, 12:22 PM   #10
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I quit!
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