Results 1 to 2 of 2

Math Help - A few more log questions

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    A few more log questions

    Hi
    Need help on the following:
    1)Make the independent variable the subject of the equation:
    u(t) = \frac{ae^t + b}{ce^t + d}

    This is what i have done, however i don't understand how the book's answer got ln(\frac{du-b}{-cu+a})

    u(ce^t + d) = ae^t +b
    uce^t + ud - ae^t - b =0
    e^t(uc - a) = -ud+b
    e^t = \frac{-(ud - b)}{uc - a}
    t=ln{-(ud - b)}{(uc - a)}

    2)Simply: \frac{(e^x)^2}{e^{2x-1}}
    Not sure how to start this.

    3)Solve for x given that: ln(x) = ln(a) +n * ln(t)
    This is what i have done:

    ln(x) = 1 + n * ln(at)
    e^{ln(x)} = e^{ln(at^{1+n}}
    x=at^{1+n}

    book's answer is at^n

    4)Solve for x given: ln(x) = ln(a) + kt
    This is what i have done:

    e^{ln(x)} = e^{ln(a)} + e^{kt}

    x = a + e^{kt}

    book's answer says its ae^{kt}

    P.S
    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 Paymemoney
    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help on the following:
    1)Make the independent variable the subject of the equation:
    u(t) = \frac{ae^t + b}{ce^t + d}

    This is what i have done, however i don't understand how the book's answer got ln(\frac{du-b}{-cu+a})

    u(ce^t + d) = ae^t +b
    uce^t + ud - ae^t - b =0
    e^t(uc - a) = -ud+b
    e^t = \frac{-(ud - b)}{uc - a}
    You are correct down to here. But what's this?
    t=ln{-(ud - b)}{(uc - a)}
    Surely you mean:
    t=\ln\left(\frac{-(ud - b)}{(uc - a)}\right)
    which is equivalent to the answer in the book. (Just multiply top-and-bottom by -1.)

    2)Simply: \frac{(e^x)^2}{e^{2x-1}}
    Not sure how to start this.
    Just use the rules of indices:
    \big(x^a\big)^b=x^{ab}
    and:
    \frac{x^a}{x^b}=x^{a-b}
    to write:
    \frac{(e^x)^2}{e^{2x-1}}=\frac{e^{2x}}{e^{2x-1}}
     =e^{2x-(2x-1)}

    =e
    3)Solve for x given that: ln(x) = ln(a) +n * ln(t)
    I'm not sure how you got this:
    ln(x) = 1 + n * ln(at)
    On the RHS, use the rule:
    n\ln (y) = \ln(y^n)
    to get:
    \ln(x) = \ln(a) +n\ln(t)

    \Rightarrow \ln(x) = \ln(a) + \ln(t^n)
    And then use:
    \ln(a) + \ln(b) = \ln(ab)
    to get:
    \Rightarrow \ln(x) = \ln(at^n)

    \Rightarrow x = at^n
    4)Solve for x given: ln(x) = ln(a) + kt
    This is what i have done:

    e^{ln(x)} = e^{ln(a)} + e^{kt}
    No. You must combine the RHS into a single logarithm:
    \ln(x) = \ln(a) + kt

    \Rightarrow \ln(x) = \ln(a) + kt\ln(e), using the fact that \ln(e) = 1.
    =\ln(a) +\ln(e^{kt})

    =\ln(ae^{kt})
    \Rightarrow x = ae{kt}
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  4. Two questions :)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 11th 2008, 01:08 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum