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Math Help - n degree function

  1. #1
    Senior Member furor celtica's Avatar
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    n degree function

    the diagram shows the graph of y=x^n, where n is an integer. given that the curve passes between the points (2,200) and (2, 2000), determine the value of n.
    i just dont get this question. what does it mean precisely to 'pass between the points', how is this interpreted algebraically? its just too random for me, any helpful hints?
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  2. #2
    Super Member Bacterius's Avatar
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    Salut,

    if y = x^n and the curve passes between A(2,200) and B(2, 2000), first note that point B is higher than point A. So it means that the curve goes under B, and over A, at x = 2. Therefore :

    200 < 2^n < 2000

    You can then take the base-2 logarithm of each member of the inequality to extract n :

    \log_2{(200)} < \log_2{(2^n)} < \log_2{(2000)}

    7.64 < n < 10.96

    So :

    7 < n < 11

    (by rounding, and noting that the question says that the curve passes between the points, not through the points).

    Therefore the only options left are \boxed{n = 8}, \boxed{n = 9} and \boxed{n = 10}

    All those values of n satisfy your statement. So how do we know which one is the good one ?

    (!) Look at the curve : the values are negative with x < 0, and positive x > 0. This is only possible if the exponent is odd (if it is even, the result is always positive). Therefore, the only solution is ... \boxed{n = 9} !

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  3. #3
    Senior Member furor celtica's Avatar
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    merci
    dis t'es pas vraiment né en 1993 non?
    rassure-moi, sinon merde j'ai la honte
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  4. #4
    Super Member Bacterius's Avatar
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    Quote Originally Posted by furor celtica View Post
    merci
    dis t'es pas vraiment né en 1993 non?
    rassure-moi, sinon merde j'ai la honte
    Désolé j'ai seize ans et demi ^^ (donc 1993 )
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  5. #5
    Senior Member furor celtica's Avatar
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    oh la la
    je me sens nul là tout a coup
    sinon
    is there any other way to get this result sans utiliser de log? parce que this is a revision question for a section in which logarithms hadnt yet been introduced. donc voilà
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  6. #6
    Super Member Bacterius's Avatar
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    I don't think so. Since the original question basically boils down to solve for an exponent, logs have to be used. Well ... so I think.
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  7. #7
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    Same question

    Hello there! I had the same question and solved it in a different way. Actually I wasn't sure if it is correct, but I got to the possibilities of n=8, n=9, n=10, and then I saw this thread.
    So, when you get to 200<2^n<2000
    you can write 25*2^3<2^n<125*2^4
    from that you have 2^4*2^3<2^n<2^7*2^4
    from which we get 2^7<2^n<2^11
    so 7<n<11. From here you get the options that n=8,9,10 (as n is an integer), and from this as 9 is the only odd one, you find the solution. I know this might be a bit late, but I just came across it!
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