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    Can someone please help me! I'm stuck on this question. Thank you in advance!

    The graph of y=ax^n passes through the points (1,2) and (-2,32)
    Find exact values of a and n
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    If y=ax^n passes through (1, 2), then you can substitute this in to find an equation. You can do the same for (-2, 32).
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    (Original post by Integer123)
    If y=ax^n passes through (1, 2), then you can substitute this in to find an equation. You can do the same for (-2, 32).
    a=2
    n=4
    ???
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    (Original post by Greenapplepear)
    a=2
    n=4
    ???
    Yeah that's correct
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    Yay, thank you so much!
    (Original post by Integer123)
    Yeah that's correct
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    jj
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    (Original post by Integer123)
    Yeah that's correct
    When you make the substitution you get:
    32 = 2 (-2)^n
    -2^n=16

    I know n=4 is the correct answer but how do you take log of a negative number?
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    (Original post by Kartheyan)
    When you make the substitution you get:
    32 = 2 (-2)^n
    -2^n=16

    I know n=4 is the correct answer but how do you take log of a negative number?
    You can't take the log of a negative number in this context. Basically, \log(-2)=\log 2 + i\pi where i=\sqrt{-1}, so it's best to ignore the log of a negative number.

    Instead, (-2)^x only has real values when x is an integer, and is only positive when x is an even integer. By inspection, we then get n=4.
 
 
 
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