Results 1 to 3 of 3

Math Help - Simplying Log Problems

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    49

    Simplying Log Problems

    Hi My Friends
    I need ur help to simply the following problem


    can u please explain the steps to solve the problem

    Thanks in Advanced
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello metallica007
    Quote Originally Posted by metallica007 View Post
    Hi My Friends
    I need ur help to simply the following problem


    can u please explain the steps to solve the problem

    Thanks in Advanced
    The definition of a log is this (and it's a bit involved!):
    The log of a number to a certain base is that power to which the base must be raised in order to get the number.
    In other words, if
    x = \log_b(y)
    then the log of y to base b is x. So when we raise b to the power x we get y. In other words:
    b^x = y
    Make sure you can get your mind around that!

    The laws of logs (whatever the base) are:
    \log (a) + \log (b) = \log(ab)... (1)

    \log(a) -\log(b) = \log\left(\frac{a}{b}\right)
    ... (2)

    a\log(b) = \log\left(b^a\right)
    ... (3)
    Now to your question. Note first that \sqrt5 = 5^{\frac12}. So we get:
    \log_5\left(\frac{\sqrt5}{5}\right) = \log_5(\sqrt5) - \log_5(5), using law (2) above
    =\log_5(5^{\frac12}) - \log_5(5^1)

    =\tfrac12-1, using the definition of a log

    =-\tfrac12
    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    Quote Originally Posted by Grandad View Post
    Hello metallica007The definition of a log is this (and it's a bit involved!):
    The log of a number to a certain base is that power to which the base must be raised in order to get the number.
    In other words, if
    x = \log_b(y)
    then the log of y to base b is x. So when we raise b to the power x we get y. In other words:
    b^x = y
    Make sure you can get your mind around that!

    The laws of logs (whatever the base) are:
    \log (a) + \log (b) = \log(ab)... (1)

    \log(a) -\log(b) = \log\left(\frac{a}{b}\right)
    ... (2)

    a\log(b) = \log\left(b^a\right)
    ... (3)
    Now to your question. Note first that \sqrt5 = 5^{\frac12}. So we get:
    \log_5\left(\frac{\sqrt5}{5}\right) = \log_5(\sqrt5) - \log_5(5), using law (2) above
    =\log_5(5^{\frac12}) - \log_5(5^1)

    =\tfrac12-1, using the definition of a log

    =-\tfrac12
    Grandad
    Great Work
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. binomial problems/normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 19th 2010, 11:46 PM
  2. simplying trig expression
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 31st 2009, 09:29 AM
  3. simplying indices
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 17th 2009, 09:17 AM
  4. simplying expressions (indices)
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 9th 2008, 09:04 AM
  5. Replies: 1
    Last Post: August 3rd 2008, 12:31 PM

Search Tags


/mathhelpforum @mathhelpforum