Results 1 to 12 of 12

Math Help - Check my work

  1. #1
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024

    Check my work

    I've enclosed an excel program I made to find all different possible human dna.
    I would like someone to check the work please.
    note: you need to click on the green tab to get to what I want checked
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024

    note

    I'm not asking you to check my data, what I am asking is if...
    [ ( 2 ^ r ) ^ g ] ^ c

    [ 2 ^ ( r g ) ] ^ c

    2 ^ ( r g c )

    2 ^ ( 5000 543.4782609 46 )

    2 ^ ( 2717391.304 46 )

    2 ^ 125000000

    ( 2 ^ 1000 ) ^ 125000

    ( 1.0715 10 ^ 301 ) ^ 125000

    1.0715 ^ 125000 10 ^ ( 301 125000 )

    ( 1.0715 ^ 1000 ) ^ 125 10 ^ 37625000

    ( 9.82144 10 ^ 29 ) ^ 125 10 ^ 37625000

    9.82144 ^ 125 10 ^ ( 29 125 ) 10 ^ 37625000

    9.82144 ^ 125 10 ^ 3625 10 ^ 37625000

    9.82144 ^ 125 10 ^ ( 3625 + 37625000 )

    9.82144 ^ 125 10 ^ 37628625

    ( 9.82144 ^ 100 ) ^ 1.25 10 ^ 37628625

    ( 1.6501 10 ^ 99 ) ^ 1.25 10 ^ 37628625

    1.6501 ^ 1.25 10 ^ ( 99 1.25 ) 10 ^ 37628625

    1.870199043 10 ^ 123.75 10 ^ 37628625

    1.870199043 10 ^ ( 123.75 + 37628625 )

    1.870199043 10 ^ 37628748.75

    is correct.

    in other words...
    does 2^{125000000}=1.870199043	\cdot	10^{37628748.75}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Quick

    in other words...
    does 2^{125000000}=1.870199043	\cdot	10^{37628748.75}
    In otherwords is:

    125000000=\log_2(1.870199043)+37628748.75\times \log_2(10)?

    Code:
    >
    >logbase(1.870199043,2)+logbase(10,2)*37628748.75
          124999998.551 
    >
    >
    Is that close enough? All the arithmetic is double precision and as all
    quantities involved are positive I expect this is good to at least 10 significant
    digits.

    Checking this in an arbitrary precisions arithmetic package tell me all the
    quoted figures above are correct (if I have typed the numerical constants
    correctly)

    RonL
    Last edited by CaptainBlack; July 23rd 2006 at 12:46 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by CaptainBlack
    In otherwords is:

    125000000=\log_2(1.870199043)+37628748.75*\log_2(1  0)?

    Code:
    >
    >logbase(1.870199043,2)+logbase(10,2)*37628748.75
          124999998.551 
    >
    >
    Is that close enough? All the arithmetic is double precision and as all
    quantities involved are positive I expect this is good to at least 10 significant
    digits.

    Checking this in an arbitrary precisions arithmetic package tell me all the
    quoted figures above are correct (if I have typed the numerical constants
    correctly)

    RonL
    How do you use logarithims, it seem to me that if you have something like 2^5=... than you would get rid of the two and multiply everything by log_2 and then you can use the rules of logarithims to say that log(x^y)=y*log(x), am I right? (I'm not really sure about the rules of Logarithims)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Quick
    How do you use logarithims, it seem to me that if you have something like 2^5=... than you would get rid of the two and multiply everything by log_2 and then you can use the rules of logarithims to say that log(x^y)=y*log(x), am I right? (I'm not really sure about the rules of Logarithims)
    if 2^x=y, then x=\log_2(y),

    and for an arbitary base:

    if b^x=y, then x=\log_b(y).

    Basic rules of logarithms can be derived from the rules for exponents:

    \log_b(x.y)=\log_b(x)+\log_b(y),

    \log_b(x^n)=n\log_b(x),

    \log_a(x)=\frac{\log_b(x)}{\log_b(a)}

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Another way to think of logarithms is of functions.
    The function y=a^x a>0 has an inverse function (I assume you know what that means).
    You define that function as,
    y=\log_a x
    -----
    An inverse function is one such as,
    f(f^{-1}(x))=f^{-1}(f(x))=x

    The interesting property between exponent and logarithms is that they express a sum as a product and a product as a sum respectively.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Another way to think of logarithms is of functions.
    The function y=a^x a>0 has an inverse function (I assume you know what that means).
    You define that function as,
    y=\log_a x
    -----
    An inverse function is one such as,
    f(f^{-1}(x))=f^{-1}(f(x))=x

    The interesting property between exponent and logarithms is that they express a sum as a product and a product as a sum respectively.
    What I find surprising is that Quick, who displays significant mathematical
    knowlege should be lacking in the logarithm department.

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by ThePerfectHacker
    Another way to think of logarithms is of functions.
    The function y=a^x a>0 has an inverse function (I assume you know what that means).
    You define that function as,
    y=\log_a x
    shouldn't it be \log_a(y)=x
    I assume you know what that means
    you assume too much! I have been trying to figure what an inverse function is ever since I got here, I've been starting to believe that the inverse of a function is a derivative of the function, which I would like to know more about...

    Quote Originally Posted by CaptainBlack
    What I find surprising is that Quick, who displays significant mathematical
    knowlege

    Quote Originally Posted by CaptainBlack
    should be lacking in the logarithm department.
    Give a break to a high-school freshman.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Quick
    shouldn't it be \log_a(y)=x
    No
    Quote Originally Posted by Quick
    what an inverse function is ever since I got here,
    When you come back I can give you an online lecture of what an inverse function is. When I was younger I also never understood it (because high schools is too informal). It makes perfect sense with sets
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    just a quick question: does \log_bb=1? (that's the impression I'm getting but I want to be sure)

    Quote Originally Posted by ThePerfectHacker
    When you come back I can give you an online lecture of what an inverse function is. When I was younger I also never understood it (because high schools is too informal). It makes perfect sense with sets
    I'm back...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Quick
    just a quick question: does \log_bb=1? (that's the impression I'm getting but I want to be sure)
    Yes
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by ThePerfectHacker
    When you come back I can give you an online lecture of what an inverse function is. When I was younger I also never understood it (because high schools is too informal). It makes perfect sense with sets
    I already know what an inverse function is, (there are many websites besides MHF) but none of the sites I've been two explain how to find it...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 12th 2011, 04:05 AM
  2. Work--please check my work for me
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 1st 2011, 09:02 AM
  3. Replies: 1
    Last Post: July 21st 2010, 06:02 PM
  4. check my work please
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 20th 2007, 07:11 PM
  5. can someone check my work thanks!!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: January 3rd 2007, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum