May 2nd, 2019, 11:52 AM  #11 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  $\displaystyle \sqrt{2^xa}\Rightarrow 2^xa\geq0\\ \\ \sqrt{2^x8}\Rightarrow 2^x\geq8\Rightarrow x\geq3$ in this case, the domain of existence of the initial equation is $\displaystyle [3, \infty)$ Last edited by aurel5; May 2nd, 2019 at 11:56 AM. 
May 2nd, 2019, 05:38 PM  #12 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
The domain doesn't have to be an interval. For $x = 2,\ \sqrt{2^x  8}$ is imaginary, but $(x^2  5x + 6)\sqrt{2^x  8}$ is still zero. 
May 2nd, 2019, 06:16 PM  #13 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  I think it's very dangerous, because x is only real !! $\displaystyle \sqrt{2^xa}\Rightarrow x\geq log_2 a$, for a>0 $\displaystyle For\ \ x = 2 \Rightarrow \sqrt{2^2a} \Rightarrow 4a\geq0 \Rightarrow a\leq 4$ We must take care of "a", but also of "x", in R (!) Last edited by aurel5; May 2nd, 2019 at 06:39 PM. 

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