Results 1 to 6 of 6

Math Help - Natural logarithm problem

  1. #1
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200

    Natural logarithm problem

    I am studying for a math competition at my school and this here just stumps me.
    \textrm{ln}\frac{3x^4}{5y^2}

    My work:
    Quotient property:

    \textrm{ln}3x^4-\textrm{ln}5y^2

    Power property:

    4\textrm{ln}3x-2\textrm{ln}5y

    Product property:


    \textrm{ln}3+4\textrm{ln}x-\textrm{ln}5+2\textrm{ln}y

    And this is what I have got after a long time of trying...
    But the answer key says the answer is:

    \textrm{ln}3+4\textrm{ln}x-\textrm{ln}5-2\textrm{ln}y
    can someone explain this to me?
    thanks.


    \int
    Last edited by integral; March 3rd 2010 at 03:02 PM. Reason: spelling error :)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie Wittgenstein's Avatar
    Joined
    Mar 2010
    Posts
    3
    \text{ln}ab-\text{ln}cd= \text{ln}a+\text{ln}b-(\text{ln}c+\text{ln}d), not \text{ln}a+\text{ln}b-\text{ln}c+\text{ln}d
    Last edited by Wittgenstein; March 3rd 2010 at 03:45 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by integral View Post
    \textrm{ln}\frac{3x^4}{5y^2}

    My work:
    Quotient property:

    \textrm{ln}3x^4-\textrm{ln}5y^2

    Power property:

    4\textrm{ln}3x-2\textrm{ln}5y
    Not sure if this is correct,

    Is is true that \ln(a)^b = b\ln(a)

    But you have \ln(3x^4) = \ln(3)+\ln(x^4) = \ln(3)+4\ln(x)

    You have somehow ended up with the correct working but only by fluke.

    What is the question actually asking?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie Wittgenstein's Avatar
    Joined
    Mar 2010
    Posts
    3
    Quote Originally Posted by pickslides View Post
    What is the question actually asking?
    I think it was asking to express \textrm{ln}\frac{3x^4}{5y^2}<br />
in terms of x and y. So we have: \textrm{ln}\frac{3x^4}{5y^2} = \textrm{ln}{3x^4}-\textrm{ln}{5y^2} = \textrm{ln}{3}+\textrm{ln}{x^4}-(\textrm{ln}{5}+\textrm{ln}{y^2}) = \textrm{ln}{3}+\textrm{4ln}{x}-(\textrm{ln}{5}+\textrm{2ln}{y}) , but the OP thought it as \textrm{ln}{3}+\textrm{4ln}{x}-\textrm{ln}{5}+\textrm{2ln}{y} instead of \textrm{ln}{3}+\textrm{4ln}{x}-\textrm{ln}{5}-\textrm{2ln}{y}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member integral's Avatar
    Joined
    Dec 2009
    From
    Arkansas
    Posts
    200
    Quote Originally Posted by pickslides View Post
    Not sure if this is correct,

    Is is true that \ln(a)^b = b\ln(a)

    But you have \ln(3x^4) = \ln(3)+\ln(x^4) = \ln(3)+4\ln(x)
    What? product rule states: log_buv=log_bu+log_bv<br />
 So if I did: ln(3x^4)=4ln(3x) it would still end up as
    \ln(3)+4\ln(x)
    You have somehow ended up with the correct working but only by fluke.

    What is the question actually asking?
    .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    I think someone above has already corrected the error. I personally think that you made a tiny mistake after the above stage. It should be
    =\textrm{ln}3+4\textrm{ln}x-(\textrm{ln}5 + \textrm{ln}y^2)
    The negative applies to both of the two log functions at the end, just as it does at the beginning of my post when they are combined.
    This then becomes the correct answer of:
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 16
    Last Post: April 11th 2010, 08:33 PM
  2. natural logarithm
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: June 15th 2009, 07:25 PM
  3. Natural logarithm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 7th 2009, 07:38 PM
  4. Natural Logarithm
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 4th 2008, 01:08 AM
  5. natural logarithm integrals
    Posted in the Calculus Forum
    Replies: 8
    Last Post: June 23rd 2007, 05:06 PM

Search Tags


/mathhelpforum @mathhelpforum