You don't have to translate it.. just looking on it: set A is given by the first inqeuality and set B is given by the second inequality, we need to find [math]A\cap B[/math]Hi S.A.M. Part of your post is not English. Can you translate? Otherwise, what have you tried, so far?
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Hi canvas. I agree with that statement!You don't have to translate it …
The translation would be just like that, thanks!You don't have to translate it.. just looking on it: set A is given by the first inqeuality and set B is given by the second inequality, we need to find [math]A\cap B[/math]
I‘ve tried and got results for both of them, but I‘m not sure if the answers are correct or not.Hi S.A.M. Part of your post is not English. Can you translate? Otherwise, what have you tried, so far?
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So what are the results you got?I‘ve tried and got results for both of them, but I‘m not sure if the answers are correct or not.
For A, I agree!I got [1/2,2] for set A and (1/2,inf) for set B
Is the number 1?For A, I agree!
For B, please check your work. Using Change-of-Base formula on term log_x(8) would yield ln(8)/ln(x). That tells us there is at least one number in your solution set that you need to remove from the domain. Can you see what it is? Think about the graph of ln(x). Also, there is a short interval missing from your solution and one interval in there that produces values less than zero. We could confirm that by using a calculator to check some x-values on either side of that discontinuity in the domain I'd mentioned above.
Regarding your posted solution's parentheses, recall that the inequality is 'greater than or equal to zero'.
If you get stuck, consider showing some of your work. Maybe the mistake is something easy to fix.
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Yes. The left-hand side of that inequality is not defined when x=1.Is the number 1?
Hi I_L. Your set B may not contain negative numbers or zero.Set A:[1/2 , 2]
Set B: ( - inf , 1/2] ∪ (1 , +inf)
Thus : A ∩ B = {1/2} ∪ (1,2]
That's okay; I'll say it another way.… I can't understand what you mean.
Thank you so much for pointing out this important point that I forgot.That's okay; I'll say it another way.
logx(a)
Symbol x is the base of the logarithm. It cannot be negative, zero or one. (Any positive Real number other than 1 is okay.)
Here's another way to see that x cannot be negative, in the given exercise. We have the logarithmic term:
log2(√[2x])
√[2x] is not defined for negative x, in the Real number system.
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