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 February 22nd, 2017, 05:44 AM #1 Newbie   Joined: May 2016 From: Việt 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 07:17 AM. February 22nd, 2017, 09:13 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,124 Thanks: 1003 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, 09:49 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, 10:00 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 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, 10:02 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,124 Thanks: 1003 Bad eyes...I read OP as 2^(2*x) Yer peepers better than mine! February 22nd, 2017, 10:06 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, 10:23 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,124 Thanks: 1003 OK; so, using 1st equation: u = log(2), v = log(4y^2+8y+4) -x = log(u/v) / u Oui? February 22nd, 2017, 11:01 AM   #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, 11:05 AM #9 Senior Member   Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276 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, 11:22 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,124 Thanks: 1003 I quit! Tags equation, equations, solve, system Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post levietdung1998 Algebra 1 September 8th, 2014 08:42 AM wannabemathlete Algebra 2 June 5th, 2014 09:05 PM roberthun Algebra 3 January 30th, 2013 04:52 AM abotaha Calculus 2 August 3rd, 2010 09:43 AM Axle12693 Algebra 2 January 22nd, 2010 07:41 PM

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