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Thread: Why is 10^lg 5=5 and e^ln5=5, etc?

  1. #1
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    Why is 10^lg 5=5 and e^ln5=5, etc?

    Why is

    $\displaystyle (10^\lg5)=5$
    $\displaystyle (e^\ln5)=5$

    so on and so forth?
    Last edited by stupidguy; Sep 7th 2010 at 05:59 AM.
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  2. #2
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    Because the notation $\displaystyle \log$ is being used to represent the logarithm of base $\displaystyle 10$, and $\displaystyle \ln$ is being used to represent the logarithm of base $\displaystyle e$.
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    Quote Originally Posted by Prove It View Post
    Because the notation $\displaystyle \log$ is being used to represent the logarithm of base $\displaystyle 10$, and $\displaystyle \ln$ is being used to represent the logarithm of base $\displaystyle e$.
    Can you break it down further so that I can digest that information?
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    Thread closed. I dunno why i have double thread. Moderator can delete for me and transfer this reply over to the other?
    Last edited by stupidguy; Sep 7th 2010 at 06:15 AM.
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  5. #5
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    You should know that exponentials undo logarithms and vice versa.

    So $\displaystyle a^{\log_a{x}} = x$ and $\displaystyle \log_a{(a^x)} = x$.


    In your first case, you're using $\displaystyle \lg$ to represent $\displaystyle \log_{10}$ and $\displaystyle \ln$ to represent $\displaystyle \log_e$.
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  6. #6
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    Quote Originally Posted by stupidguy View Post
    Thread closed. I dunno why i have double thread. Moderator can delete for me and transfer this reply over to the other?
    Duplicate thread: http://www.mathhelpforum.com/math-he...tc-155461.html

    Thread closed.
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