My Math Forum How could I solve this system of equations???

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 February 22nd, 2017, 06:44 AM #1 Newbie   Joined: May 2016 From: Viá»‡t Nam Posts: 8 Thanks: 1 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.
 February 22nd, 2017, 10:13 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,597 Thanks: 954 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...
February 22nd, 2017, 10:49 AM   #3
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Quote:
 Originally Posted by Denis 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

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

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

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

$2(x + 1) = 2^{64} \implies x = 2^{63} - 1 \ne 3.$

 February 22nd, 2017, 11:00 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,957 Thanks: 1844 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.
 February 22nd, 2017, 11:02 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,597 Thanks: 954 Bad eyes...I read OP as 2^(2*x) Yer peepers better than mine!
February 22nd, 2017, 11:06 AM   #6
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Quote:
 Originally Posted by Denis 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

$x=y \approx 2.49$

$x=y \approx -0.343$

 February 22nd, 2017, 11:23 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,597 Thanks: 954 OK; so, using 1st equation: u = log(2), v = log(4y^2+8y+4) -x = log(u/v) / u Oui?
February 22nd, 2017, 12:01 PM   #8
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Quote:
 Originally Posted by Denis 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 $x=y$ !

 February 22nd, 2017, 12:05 PM #9 Senior Member     Joined: Sep 2015 From: USA Posts: 2,199 Thanks: 1153 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.
 February 22nd, 2017, 12:22 PM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,597 Thanks: 954 I quit!

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