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Math Help - adding exponential logs

  1. #1
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    adding exponential logs

    Hi, i would like to ask you for some help about how to add these exponential logs...

    If A = 200e^-0.2m
    and B = 10.5e^0.3m

    what is A + B

    seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...
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  2. #2
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    Quote Originally Posted by popsterrockz View Post
    Hi, i would like to ask you for some help about how to add these exponential logs...

    If A = 200e^-0.2m
    and B = 10.5e^0.3m

    what is A + B

    seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...

    There's no closed form for adding numbers raised to some power. The only thing you can do, and not always the result is nice, is to factor out:

    200\,e^{-0.2m}+10.5\,e^{0.3m}=e^{-0.2m}\left(200+10.5\,e^{0.5m}\right)

    Tonio
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  3. #3
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    exponential logs

    Quote Originally Posted by popsterrockz View Post
    Hi, i would like to ask you for some help about how to add these exponential logs...

    If A = 200e^-0.2m
    and B = 10.5e^0.3m

    what is A + B

    seriously, i cant add them... and what i mean is adding the 2 equations... and by the way i dont know how you guys can make the powers shown here so i just typed '^'... sorry...
    Hi pops,

    I did this as follows
    Take the natural log of A and its equality.Simplify

    Same procedure ln B.

    I get ln A=-1.26m ln B =m

    A+B=e^-1.26 +e^m


    bjh
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    Quote Originally Posted by bjhopper View Post
    Hi pops,

    I did this as follows
    Take the natural log of A and its equality.Simplify

    Same procedure ln B.

    I get ln A=-1.26m ln B =m

    A+B=e^-1.26 +e^m


    bjh

    The above is wrong: it is not true that \log (A+B)=\log A+\log B .

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    The above is wrong: it is not true that \log (A+B)=\log A+\log B .

    Tonio
    Hello Tonio,

    You may find an error in what I did but i am not adding the ln A +ln B. Iam adding the calculated values of A and B.If you would try this approach I would appreciate knowing where I went wrong.


    bjh
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  6. #6
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    adding exponential logs

    Quote Originally Posted by tonio View Post
    The above is wrong: it is not true that \log (A+B)=\log A+\log B .

    Tonio
    I did err in the simplification operations.Here is my revised solution.

    A=200 e^-.2m
    ln A = ln 200 e^-.2m
    ln A =ln 200 -.2m ln e
    ln A =5.3 -.2m
    e^5.3-.2m =A

    B=10.5e^.3m
    Ln B = ln 10.5 e^.3m
    ln B = ln 10.5 + .3m ln e
    lnB =2.35+.3m
    e^2.35+.3m =B

    A+B= e^5.3-.2m + e^2.35 +.3m


    bjh







    ln A =ln 200-.2m ln e
    ln A =5.3-.2m
    e^5.3-.2m =A
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  7. #7
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    [quote=bjhopper;504191]I did err in the simplification operations.Here is my revised solution.

    A=200 e^-.2m
    ln A = ln 200 e^-.2m
    ln A =ln 200 -.2m ln e
    ln A =5.3 -.2m
    e^5.3-.2m =A

    This must be A=e^{5.3-0.2m}

    B=10.5e^.3m
    Ln B = ln 10.5 e^.3m
    ln B = ln 10.5 + .3m ln e
    lnB =2.35+.3m
    e^2.35+.3m =B


    This must be B=e^{2.35+0.3m}


    A+B= e^5.3-.2m + e^2.35 +.3m


    It must be A+B=e^{5.3-0.2m}+e^{2.35+0.3m} , which is close to what I wrote in my first post (I just didn't do approximations)

    Tonio


    bjh
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  8. #8
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    Quote Originally Posted by bjhopper View Post
    Hello Tonio,

    You may find an error in what I did but i am not adding the ln A +ln B. Iam adding the calculated values of A and B.If you would try this approach I would appreciate knowing where I went wrong.


    True, you didn't do that. I got confused because your calculations are pretty wrong, but you already fixed that in other post.

    Tonio



    bjh
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