Results 1 to 3 of 3

Math Help - Question about logs

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    Question about logs

    I am studying for an end of year exam and I completely forgot everything about logs and log functions (except for the extreme basics) so can someone tell me why the answer is D by showing me all the steps?

    Any help would be greatly appreciated!
    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Question about logs-examquestionsimhavingtroublewith-2.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    May 2009
    Posts
    471
    10log a=log(a^{10})
    -3log b= -log(b^3)
    \frac{1}{2}log(9)=log(9^{\frac{1}{2}})=log \sqrt{9}=log(3)

    So...

    10log(a)-3log(b)+\frac{1}{2}log(9)=log(a^{10})-log(b^3)+log(3)

    And 2 of the log rules are:

    log(a)+log(b)=log(ab) and log(c)-log(d)=log\left(\frac{c}{d}\right)

    So...

    10log(a)-3log(b)+\frac{1}{2}log(9)=log(a^{10})-log(b^3)+log(3)=log\left(\frac{a^{10}}{b^3}\right)  +log(3)

    =log\left(\frac{3a^{10}}{b^3}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    From
    Zagreb
    Posts
    65
    log( x*y ) = log(x) + log(y)

    log( x/y ) = log(x) - log(y)

    log( x^n ) = n*log(x)

    n*log(x) = log( x^n )

    so...

    log( 3*(a^10) / b^3 ) = log( 3*( a^10 ) ) - log( b^3 ) =

    = log( 3 ) + log( a ^10 ) - 3 log( b ) =

    = 2*( 1 / 2 )*log( 3 ) + 10log( a ) - 3log( b ) =

    = (1/2)*log( 3^2 ) + 10log( a ) - 3log( b ) =

    = (1/2)*log( 9 ) + 10log( a ) - 3log( b )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question on Logs
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 1st 2009, 04:10 PM
  2. question regarding logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 1st 2009, 07:28 AM
  3. logs question
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 5th 2009, 04:13 PM
  4. One question on logs.
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 31st 2009, 12:31 PM
  5. one more question on logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 28th 2009, 10:27 PM

Search Tags


/mathhelpforum @mathhelpforum