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Thread: [SOLVED] putting a exponential growth function to log form

  1. April 7th 2010, 06:33 PM #1
    kelvinly
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    [SOLVED] putting a exponential growth function to log form

    p(t) = 100*(2^2t)

    would i have to multiply the 100 and the 2 or just leave it like that? i'm confusing myself with the logarithmic law..

    edit: thank you integral!
    Last edited by kelvinly; April 7th 2010 at 07:17 PM.
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  2. April 7th 2010, 07:16 PM #2
    integral
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    P(t)=100(2^{2t}) \Rightarrow \frac{P(t)}{100}=2^{2t}
    thus:

    log_2\frac{P(t)}{100}=2t \,\,\,\because \,\,\,a^x=b\Rightarrow log_ab=x

    then:

    \frac{1}{2}log_2\frac{P(t)}{100}=t

    log_2\sqrt{\frac{P(t)}{100}}=t

    <br /> log_2\frac{\sqrt{P(t)}}{10}=t
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  3. April 7th 2010, 07:17 PM #3
    Sudharaka
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    Dear integral,

    There is a little typo in the first line. It should be, P(t)=100(2^{2t}) \Rightarrow \frac{P(t)}{100}=2^{2t}
    Last edited by Sudharaka; April 8th 2010 at 05:44 AM.
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  4. April 7th 2010, 07:23 PM #4
    integral
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    I really meant 2^{2t}, but thank you
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  5. April 7th 2010, 07:24 PM #5
    kelvinly
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    Quote Originally Posted by integral View Post
    I really meant 2^{2t}, but thank you
    yes i know haha. no, thank you!
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  6. April 8th 2010, 03:23 AM #6
    HallsofIvy
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    Quote Originally Posted by Sudharaka View Post
    Dear integral,

    There is a little typo in the first line. It should be, P(t)=100(2^{2t}) \Rightarrow \frac{P(t)}{100}=2^{2}2^{t}
    No. 2^22^t= 2^{2+ t}, not 2^{2t}= (2^2)^t= 4^t.
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