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Math Help - Exponents

  1. #1
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    Exponents

    1. Given that 2^a = 3^b = 18^c, Show that ab=c(b+2a).

    2. Given that p^1/2 - p^(-1/2) = 2, Show that p + p^(-1) = 6
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Could you tell us what you came up with so far?

    But for the first part, I don't understand where the c comes from...
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  3. #3
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    Yeah sure....

    For Question 2,

    p^1/2 - p^(-1/2) = 2

    √p - 1/√p = 2

    Let y = √p

    y - 1/y = 2

    y^2 - 2y - 1 = 0

    I can do this question up to here....Then,I'm blurr......
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    {p^1/2 - p^(-1/2)}^2=2^2
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  5. #5
    MHF Contributor Unknown008's Avatar
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    For the second one, square both sides.

    For the first one, are you sure this is the right question?
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Sorry Unknown008
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  7. #7
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    The 1st question is the right one.....I missed the 'c' earlier....sorry!
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    18^c=(2*3^2)^c=2^c*3^2c
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  9. #9
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    For the second one,

    I'm not sure what to do after squaring it.
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    just "play" with the equations... and try to enjoy by doing it...
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  11. #11
    MHF Contributor Unknown008's Avatar
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    You got:

    (\sqrt{p} - \dfrac{1}{\sqrt{p}})^2 = 2^2

    This becomes:

    \sqrt{p}^2 - 2(\dfrac{1}{\sqrt{p}})(\sqrt{p}) + (\dfrac{1}{\sqrt{p}})^2 = 4

    p - 2 + \dfrac{1}{p} = 4
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  12. #12
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    Okay,I managed to get until p^2 - 6p + 1 = 0 ......Then how to show that p + p^(-1) = 6
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  13. #13
    MHF Contributor Unknown008's Avatar
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    From where I left, just move the '-2' and what do you get?
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  14. #14
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    Okay got it.....Thanks!!
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  15. #15
    MHF Contributor Unknown008's Avatar
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    Okay, for the next one, I'm not sure if this is the right method, but it's quite long.

    2^a = 3^b = 18^c

    2^a = 3^b = 2^c3^{2c}

    ln(2^a) = ln(3^b) = ln(2^c3^{2c})

    ln(2^a) = ln(3^b) = ln(2^c) + ln(3^{2c})

    a ln2 = bln3 = c ln2 + 2c ln3

    [From here, we know that \dfrac{ln3}{ln2} = \dfrac{a}{b}]

    Divide by ln(2)

    a  = b\dfrac{ln3}{ln2} = c  + 2c \dfrac{ln3}{ln2}

    We consider only the centre and right parts now and used what we found a little earlier to remove the natural logs.

    b(\dfrac{a}{b}) = c  + 2c (\dfrac{a}{b})

    Can you continue now?
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