Re: solve for something else

Quote:

Originally Posted by

**leviriven** 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:

to #2:

Now take the natural logarithm on both sides:

Re: solve for something else

Quote:

Originally Posted by

**leviriven** 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) $k=\dfrac{\log(A)-\log(V)}{d}$

Re: solve for something else

Quote:

Originally Posted by

**leviriven** 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.**

Re: solve for something else

Thank you very much! This looks great!