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Math Help - solve for something else

  1. #1
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    Post solve for something else

    I frequently use the following two formulas in my work.

    (1) V = A / (1 + kd)
    (2) V = Ae^(-kd)

    These formulas provide alternative methods for finding the value V, when the variables A, k, and d are known. However, currently I have the values for V, d, and A, but not for k. I would like to reformulate these expressions to solve for k.

    I have completely forgotten the necessary algebraic rules. I would greatly appreciate if someone can help me reformulate these equations.
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  2. #2
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    Re: solve for something else

    Quote Originally Posted by leviriven View Post
    I frequently use the following two formulas in my work.

    (1) V = A / (1 + kd)
    (2) V = Ae^(-kd)

    These formulas provide alternative methods for finding the value V, when the variables A, k, and d are known. However, currently I have the values for V, d, and A, but not for k. I would like to reformulate these expressions to solve for k.

    I have completely forgotten the necessary algebraic rules. I would greatly appreciate if someone can help me reformulate these equations.
    Helo,

    to #1:

    V = \frac{A }{1 + k\cdot d}~\implies~V(1+k\cdot d)=A~\implies~V+V \cdot d \cdot k = A\\ \\ \implies~V \cdot d \cdot k = A-V~\implies~k=\frac{A-V}{V \cdot d }

    to #2:

    V = A \cdot e^{-k\cdot d}~\implies~\frac VA=e^{-k\cdot d} Now take the natural logarithm on both sides:

    \ln\left(\frac VA \right)= - k \cdot d~\implies~k = -\frac1d \cdot (\ln(V) - \ln(A))
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  3. #3
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    Re: solve for something else

    Quote Originally Posted by leviriven View Post
    I frequently use the following two formulas in my work.
    (1) V = A / (1 + kd)
    (2) V = Ae^(-kd).
    (1) $k=\dfrac{A-V}{dV}$

    Assume all variables are positive and use the natural logarithm:

    (2) $k=\dfrac{\log(A)-\log(V)}{d}$
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  4. #4
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    Re: solve for something else

    Quote Originally Posted by leviriven View Post
    I frequently use the following two formulas in my work.

    (1) V = A / (1 + kd)
    (2) V = Ae^(-kd)

    These formulas provide alternative methods for finding the value V, when the variables A, k, and d are known. However, currently I have the values for V, d, and A, but not for k. I would like to reformulate these expressions to solve for k.

    I have completely forgotten the necessary algebraic rules. I would greatly appreciate if someone can help me reformulate these equations.
    Equation I: $d \ne 0\ and\ V = \dfrac{A}{1 + kd} \implies V + kdV = A \implies kdV = A - V \implies k = \dfrac{A - V}{dV}.$

    In the next equation, I am assuming that e is Euler's number, a constant, rather than a variable.

    Equation II: $d \ne 0\ and\ V = Ae^{(-kd)} \implies ln(V) = ln\left(Ae^{(-kd)}\right) = ln(A) - kd *ln(e) = ln(A) * kd \implies$

    $kd = ln(A) - ln(V)= ln\left(\dfrac{A}{V}\right) \implies k = \dfrac{1}{d} * ln\left(\dfrac{A}{V}\right) =ln\left(\left\{\dfrac{A}{V}\right\}^{(1/d)}\right) \implies k = ln\left(\sqrt[d]{\dfrac{A}{V}}\right).$

    EDIT: I see others posted while I was writing. The three results for Equation II compute to the same answer.
    Last edited by JeffM; April 15th 2014 at 12:09 PM.
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  5. #5
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    Re: solve for something else

    Thank you very much! This looks great!
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