Results 1 to 5 of 5

Thread: Logarithm

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    145

    Logarithm

    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    14,797
    Thanks
    3013
    Quote Originally Posted by Hellbent View Post
    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012
    logs could be used ... but the way you did it more direct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    145
    I need assistance with the logs part, please.
    Tried using logs prior to posting question, but the answers are inconsistent.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    584
    Quote Originally Posted by Hellbent View Post
    Hi,

    Are there alternative ways of solving this equation? Could logarithms be used?

    a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012
    60\log{a} = \log{2.044}

    \log{a} = \frac{log{2.044}}{60}<br />
\RightArrow<br />
= .00517<br />

    so 10^{(.00517)} = 1.01199 or 1.012

    as mentioned your first method is better..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    14,797
    Thanks
    3013
    a^{60} = k , where k = 2.044

    60\log{a} = \log{k}<br />

    \displaystyle \log{a} = \frac{\log{k}}{60}

    \displaystyle a = e^{\frac{\log{k}}{60}}

    a \approx 1.012

    note that \displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Jul 18th 2011, 07:04 AM
  2. logarithm
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jul 17th 2010, 04:23 AM
  3. Logarithm #2
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jun 1st 2010, 05:29 AM
  4. Logarithm
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Sep 17th 2009, 10:23 PM
  5. Logarithm
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Mar 13th 2008, 10:05 AM

Search Tags


/mathhelpforum @mathhelpforum