Results 1 to 2 of 2

Thread: Logarithms

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    9

    Logarithms

    Thanks for your help in advance..

    If $\displaystyle a = \log_{10} \frac{10}{9} , b = \log_{10} \frac{25}{24} and
    c = \log_{10} \frac{81}{80} $

    prove that $\displaystyle \log_{10} 2 = 7a-2b+3c $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by ecogreen View Post
    Thanks for your help in advance..

    If $\displaystyle a = \log_{10} \frac{10}{9} , b = \log_{10} \frac{25}{24} and
    c = \log_{10} \frac{81}{80} $

    prove that $\displaystyle \log_{10} 2 = 7a-2b+3c $

    $\displaystyle 7a-2b+3c=7\log\frac{10}{9}-2\log\frac{25}{24}+3\log\frac{81}{80}$ $\displaystyle =7(\log 2+\log 5-2\log 3)-2(2\log 5-3\log 2-\log 3)+3(4\log 3-4\log 2 -\log 5)$ , and now open up parenthese and you're done.

    Logarithms' properties used: $\displaystyle \log\frac{A}{B}=\log A-\log B\,,\,\log(AB)=\log A+\log B\,,\,\log(A^n)=n\log A$

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logarithms help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 6th 2010, 06:29 PM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 6th 2010, 04:46 PM
  3. logarithms
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 24th 2010, 04:26 AM
  4. Logarithms
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 18th 2010, 02:52 PM
  5. logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 16th 2008, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum