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May 2nd, 2019, 10:52 AM   #11
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Quote:
 Originally Posted by skipjack Why?
$\displaystyle \sqrt{2^x-a}\Rightarrow 2^x-a\geq0\\ \\ \sqrt{2^x-8}\Rightarrow 2^x\geq8\Rightarrow x\geq3$

in this case, the domain of existence of the initial equation is $\displaystyle [3, \infty)$

Last edited by aurel5; May 2nd, 2019 at 10:56 AM.

 May 2nd, 2019, 04:38 PM #12 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 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, 05:16 PM   #13
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Quote:
 Originally Posted by skipjack For $x = 2,\ \sqrt{2^x - 8}$ is imaginary

I think it's very dangerous, because x is only real !!

$\displaystyle \sqrt{2^x-a}\Rightarrow x\geq log_2 a$, for a>0

$\displaystyle For\ \ x = 2 \Rightarrow \sqrt{2^2-a} \Rightarrow 4-a\geq0 \Rightarrow a\leq 4$

We must take care of "a", but also of "x", in R (!)

Last edited by aurel5; May 2nd, 2019 at 05:39 PM.

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