- #1

- 24

- 0

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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- Thread starter impendingChaos
- Start date

- #1

- 24

- 0

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

- #2

- 53

- 0

There's usually no nice simplification when there's a sum in the denominator.

- #3

- 1,425

- 1

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:

[tex]\frac{Ae^{Cx} + B}{Oe^{Px} + Q}[/tex]

For example, let's say A = 1 and P = 1

[tex]\frac{Ae^{x} + B}{Oe^{x} + Q}[/tex]

Make the substitution e^x = y and get

[tex]\frac{Ay + B}{Oy + Q}[/tex]

Now we can write

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

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

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

B - QA/O is the remainder here.

[tex]\frac{Ae^{Cx} + B}{Oe^{Px} + Q}[/tex]

For example, let's say A = 1 and P = 1

[tex]\frac{Ae^{x} + B}{Oe^{x} + Q}[/tex]

Make the substitution e^x = y and get

[tex]\frac{Ay + B}{Oy + Q}[/tex]

Now we can write

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

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

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

B - QA/O is the remainder here.

Last edited:

- #4

- 24

- 0

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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