Page 2 of 2 FirstFirst 12
Results 16 to 24 of 24

Math Help - logarithms

  1. #16
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BlueStar View Post
    ...
    e^3x = 6 (divide the other 3?)
    ...
    Here is your problem.

    the 3 on the left is a power, how did you manage to divide by it?!
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    What do I do then? That's what I'm asking.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BlueStar View Post
    What do I do then? That's what I'm asking.
    take the log of both sides and bring the power down.

    e^{3x} = 6

    \Rightarrow \ln e^{3x} = \ln 6

    \Rightarrow 3x \ln e = \ln 6

    \Rightarrow 3x = \ln 6

    \Rightarrow x = \frac { \ln 6}{3}


    This is my 27th post!!!!
    Last edited by Jhevon; July 7th 2007 at 11:03 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Thanks!! That's been bugging me for a while.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Now for application:


    Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

    So y = 1e^0.0856t (1 is the whole amount of the substance)

    Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BlueStar View Post
    Now for application:


    Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

    So y = 1e^0.0856t (1 is the whole amount of the substance)

    Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?
    the long, hard systematic way to do this is assume you start with 1 unit, then after the half-life you would end up with 1/2 units, so you simply solve:

    \frac {1}{2} = e^{-0.0856t} for t

    the easier way is to know that for the decay equation in the form y = A_0 e^{-rt} the half-life is given by:

    t_h = \frac { \ln 2}{r}
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,079
    Thanks
    375
    Awards
    1
    Quote Originally Posted by BlueStar View Post
    Now for application:


    Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

    So y = 1e^0.0856t (1 is the whole amount of the substance)

    Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?
    First, for a new question create a new thread.

    Second, please get in the habit of using parenthesis for these. The function you want is
    y = 1e^(-0.0856t)
    (You left out the minus sign as well!)

    Okay, now to your question. You want the "half-life" so you wish to know for what t value is y = (1/2)a.

    \frac{1}{2}a =  ae^{-0.0856t}

    \frac{1}{2} =  e^{-0.0856t}

    ln \left ( \frac{1}{2} \right ) = -0.0856t

    Now, ln \left ( \frac{1}{2} \right ) = ln(2^{-1}) = -ln(2) so...

    -ln(2) = -0.0856t

    t = \frac{ln(2)}{0.0856} \approx 8.09751

    where t is in whatever time unit you are using, in this case days.

    -Dan

    PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.
    i agree, especially for students who have little experience with the material. the manipulation is important for them to learn.

    there are a lot of things in math i usually do the hard way as opposed to memorizing formulas (for some reason i find it easier to learn how to work something out than memorize a formula), but this isn't one of them
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Quote Originally Posted by topsquark View Post
    First, for a new question create a new thread.
    Sorry.

    Okay, now to your question. You want the "half-life" so you wish to know for what t value is y = (1/2)a.

    \frac{1}{2}a = ae^{-0.0856t}

    \frac{1}{2} = e^{-0.0856t}

    ln \left ( \frac{1}{2} \right ) = -0.0856t

    Now, ln \left ( \frac{1}{2} \right ) = ln(2^{-1}) = -ln(2) so...

    -ln(2) = -0.0856t

    t = \frac{ln(2)}{0.0856} \approx 8.09751

    where t is in whatever time unit you are using, in this case days.

    -Dan

    PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.
    Doesn't seem any harder or longer than any other equations. In fact I get that one better. Thanks guys.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

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

Search Tags


/mathhelpforum @mathhelpforum