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Thread: A few more log questions

  1. #1
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    A few more log questions

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

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

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

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

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

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

    book's answer is $\displaystyle at^n$

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

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

    $\displaystyle x = a + e^{kt}$

    book's answer says its $\displaystyle ae^{kt}$

    P.S
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  2. #2
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    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:
    $\displaystyle u(t) = \frac{ae^t + b}{ce^t + d}$

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

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

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

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

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

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

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

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

    $\displaystyle =\ln(ae^{kt})$
    $\displaystyle \Rightarrow x = ae{kt}$
    Grandad
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