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Log/Index question which I am stumped with.

  1. Nov 20, 2009 #1
    I had this question in the problem solving part of my Maths Exam. When I extrapolated all the critical information from the question, I narrowed it down to this equation. However, I just couldn't solve the equation. Could anyone please help? This has been bugging me for the whole day.


    y = 2000 x 0.95^x - equation 1
    y = 3000 x 0.90^x - equation 2
     
  2. jcsd
  3. Nov 20, 2009 #2

    tiny-tim

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    Hi Procrastinate! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    So 0.95x = 3/2 0.90x

    ok, now take logs. :smile:
     
  4. Nov 20, 2009 #3
    Wouldn't that be log 0.95 3/20.9x = x

    That's what I did, but the fact that there was an x still in the index was what hindered me from completing it.
     
  5. Nov 20, 2009 #4

    tiny-tim

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    ah … you need this equation …

    keeping to the same base, a, what's loga(b2) ?

    and what's loga(bc) ? :smile:
     
  6. Nov 20, 2009 #5
    Sorry, but I am still a bit confused about the notation. Do I square it first and then do loga(bc)?
     
  7. Nov 20, 2009 #6

    tiny-tim

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    They're two separate questions …

    I thought I'd start you off with an easy one first …

    what's loga(b2) ?
     
  8. Nov 20, 2009 #7
    2logab
     
  9. Nov 21, 2009 #8

    tiny-tim

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    (just got up :zzz: …)

    Yup! :smile:

    Now what's loga(bc) ? :wink:
     
  10. Nov 21, 2009 #9
  11. Nov 21, 2009 #10

    tiny-tim

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    Yes!

    So if you take logs of 0.95x = 3/2 0.90x , you get … ? :smile:
     
  12. Nov 21, 2009 #11
    xlog0.953/2x0.9 = xlog0.92/3x0.95
     
  13. Nov 21, 2009 #12

    tiny-tim

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    I'm not following that at all :confused:

    This is supposed to be taking logs of 0.95x = 3/2 0.90x

    take them in the same base (preferably 10 :wink:).
     
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