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  
Senior Member Joined: May 2016 From: USA Posts: 1,206 Thanks: 494  Quote:
$x = 3 \implies \sqrt{2^{2^3}} = \sqrt{2^8} = 2^4 = 16 \implies$ $2(y + 1) = 16 \implies y + 1 = 8 \implies y = 7 \ne 3.$ $y = 7 \implies \sqrt{2^{2^7}} = \sqrt{2^{128}} = 2^{64}.$ $2(x + 1) = 2^{64} \implies x = 2^{63}  1 \ne 3.$  
February 22nd, 2017, 11:00 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,957 Thanks: 1844 
If the righthand 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,199 Thanks: 1153  Quote:
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 
Senior Member Joined: Sep 2015 From: USA Posts: 2,199 Thanks: 1153  
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 $\large 2=(x+1)^{\frac{1}{2^{x1}1}}$ 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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