Dividing Exponential Functions

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Dividing two exponential functions can be approached by substituting e^x with a variable, such as y, to simplify the expression. For example, when dividing functions of the form A*e^(Cx) + B by O*e^(Px) + Q, the expression can be rewritten as (Ay + B)/(Oy + Q). This method allows for the identification of a remainder, which may not yield a straightforward simplification. Additionally, converting the functions into linear expressions using natural logarithms may be complicated due to logarithmic rules. The goal is to establish a relationship between the variables n and v, which requires further exploration and assistance.
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How would one go about dividing two exponential functions.
Basically I have f(x)=k*g(x)
So to solve for k, k=f(x)/g(x)
How would one accomplish this when the functions are both within the format:
A*e^(Cx)+B

Thanks
 
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There's usually no nice simplification when there's a sum in the denominator.
 
There is a way to simplify, but it's not necessarily what you are looking for. You might or might not end up with an invariant remainder. Here:

\frac{Ae^{Cx} + B}{Oe^{Px} + Q}

For example, let's say A = 1 and P = 1\frac{Ae^{x} + B}{Oe^{x} + Q}

Make the substitution e^x = y and get

\frac{Ay + B}{Oy + Q}

Now we can write

\frac{A/O(Oy + Q - Q) + B}{Oy + Q}

\frac{A/O(Oy + Q ) - QA/O + B}{Oy + Q}

A/O + \frac{B - QA/O}{Oy + Q}

B - QA/O is the remainder here.
 
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Hmm I will try your method Werg but like you say I am not sure it is what I am looking for.
I am also trying to make both individual functions into linear expressions by taking the natural log of both sides however I run into natrual log rules which keep this from succeeding. Any ideas in the department?
Another note, I would like to clarify that I should have represented the functions as something like:
f(n)
g(v)
they are both describing different attributes of a system. What I am trying to accomplish is finding a relationship between n and v. Still working towards a solution so any help greatly appreciated.

C.N.
 
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