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Math Help - [SOLVED] Natural Base e

  1. #1
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    [SOLVED] Natural Base e

    I know that e can be approximated by the form 1 + \frac {1}{1!} + \frac {2}{2!} + \frac {3}{3!} + \frac {4}{4!} +... \frac {n}{n!} .

    Would \frac {2^1}{1!} + \frac {2^2}{2!} + \frac {2^3}{3!} + \frac {2^4}{4!} +... \frac {2^n}{n!} = e^2 ?

    And how would you determine an aproximate value for \frac {a^1}{1!} + \frac {a^2}{2!} + \frac {a^3}{3!} + \frac {a^4}{4!} +... \frac {a^n}{n!} ?
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  2. #2
    o_O
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    Yep:

    e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{n}}{n!} + ...

    If x = 2, you get what you get:
    e^{2} = 1+ 2 + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...

    If you have x = a, you get ... ?
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  3. #3
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    Quote Originally Posted by o_O View Post
    Yep:

    e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{n}}{n!} + ...

    If x = 2, you get what you get:
    e^{2} = 1+ 2 + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...

    If you have x = a, you get ... ?
    e^a right?

    Edit - Two more thing. I know that this might be off topic...but since you answered my other thread too...and I don't want to make another thread to ask 2 simple questions....

    What values of x would make \log x>\ln x and how?

    \log_3 8 + \log_3 y = 4\log_3 (2x) . If I wanted to solve this for "y" and I did, would this be right... y=2x^4 after condensing it and getting rid of log?
    Last edited by chrozer; April 12th 2008 at 10:44 PM.
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  4. #4
    Moo
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    Hello,

    Quote Originally Posted by chrozer View Post
    e^a right?
    Yep

    What values of x would make \log x>\ln x and how?
    If \log=\ln_{10} then \log x = \frac{\ln x}{\ln{10}}
    If inverted, then...just invert it ^^'
    Assuming that \ln{10}>1 (or \log{10}), can you conclude ?

    \log_3 8 + \log_3 y = 4\log_3 (2x) . If I wanted to solve this for "y" and I did, would this be right... y=2x^4 after condensing it and getting rid of log?
    I find the same ^^'
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  5. #5
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    Quote Originally Posted by Moo View Post
    Hello,



    Yep



    If \log=\ln_{10} then \log x = \frac{\ln x}{\ln{10}}
    If inverted, then...just invert it ^^'
    Assuming that \ln{10}>1 (or \log{10}), can you conclude ?



    I find the same ^^'
    Sorry but I have no idea how to conclude that.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    You have

    Quote Originally Posted by chrozer View Post
    Sorry but I have no idea how to conclude that.
    log_{10}(x)>ln(x)\Rightarrow{\frac{ln(x)}{ln(10)}>  ln(x)}=0\Rightarrow{ln(x)\bigg(\cdot{\frac{1}{ln(1  0)}-1}\bigg)>0} then using logic we see that log_{10}(x)>ln(x),\forall{x}\in(0,1)
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    log_{10}(x)>ln(x)\Rightarrow{\frac{ln(x)}{ln(10)}>  ln(x)}=0\Rightarrow{ln(x)\bigg(\cdot{\frac{1}{ln(1  0)}-1}\bigg)>0} then using logic we see that log_{10}(x)>ln(x),\forall{x}\in(0,1)
    Ok....thnx alot.
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  8. #8
    Super Member angel.white's Avatar
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    Quote Originally Posted by chrozer View Post
    \log_3 8 + \log_3 y = 4\log_3 (2x) . If I wanted to solve this for "y" and I did, would this be right... y=2x^4 after condensing it and getting rid of log?
    \log_3 8 + \log_3 y = 4\log_3 (2x)

    \log_3 (8y) = \log_3 (2x)^4

    8y = 16x^4

    y = 2x^4
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