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Math Help - Help Needed Understanding Logs

  1. #1
    Newbie bradm's Avatar
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    Help Needed Understanding Logs

    I am requesting help understanding natural logs. I understand that if you have lets say

    ln a = b

    It can be rewritten as

    a = e^b

    What I would like to know is if there is any way I can simply a function with natural logs in it,

    So if I had any function lets say

    ln x = 10 - .5 ln A + .75 ln B - 1 ln C

    Is there any way I could simplify that into normal terms without a long confusing jumble? If I had inputs, how do I go about solving a function like the above?

    My understanding of natural logs is limited.

    Thanks for any help.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    go to the link and learn the laws of logarithms ... natural and otherwise.

    Logarithms - Topics in precalculus
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bradm View Post
    I am requesting help understanding natural logs. I understand that if you have lets say

    ln a = b

    It can be rewritten as

    a = e^b

    What I would like to know is if there is any way I can simply a function with natural logs in it,

    So if I had any function lets say

    ln x = 10 - .5 ln A + .75 ln B - 1 ln C

    Is there any way I could simplify that into normal terms without a long confusing jumble? If I had inputs, how do I go about solving a function like the above?

    My understanding of natural logs is limited.

    Thanks for any help.
    Yes, you can simplify expressions like these using the laws of logarithms.

    (i) \dsiplaystyle \log_a b = c \implies a^c = b

    (ii) \dsiplaystyle n \log_a X = \log_a (X^n)

    (iii) \dsiplaystyle \log_a (XY) = \log_a X + \log_a Y

    (iv) \dsiplaystyle \log_a \frac XY = \log_a X - \log_a Y

    (v) \dsiplaystyle \log_a b = \frac {\log_c b}{\log_c a}

    Just a few of the rules that govern logs--the most important ones in my experience. Only some of these are needed to solve the problem you mentioned.
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