MHB Linearization of this equation / Inverse function

[guiismiti]

Hello,

I need to find the inverse function of the following equation

y = a * ((exp(-b * x)) + (c * (1 - (exp(-b * x)))))

Where a, b and c are constants.

I have experimental points that fit to this equation and I want to use these values in the inverse funtion to linearize it.

I have tried to use a few tools available online, but the output functions did not work, which made me think if it is actually possible to do it.Can anybody help me?
Thanks in advance.

Edited: solved

x = (1 / (-b)) * (LN(((a * c) - y) / (a * (c - 1))))

 

[MarkFL]

We are given:

$$f(x)=ae^{-bx}+c\left(1-e^{-bx}\right)=(a-c)e^{-bx}+c$$

To find the inverse function, we can write:

$$x=(a-c)e^{-by}+c$$

Solve for $y$:

$$x-c=(a-c)e^{-by}$$

$$\frac{x-c}{a-c}=e^{-by}$$

$$\ln\left(\frac{x-c}{a-c}\right)=-by$$

$$y=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)$$

Thus, we may claim:

$$f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)$$

 

[guiismiti]

We are given:

$$f(x)=ae^{-bx}+c\left(1-e^{-bx}\right)=(a-c)e^{-bx}+c$$

To find the inverse function, we can write:

$$x=(a-c)e^{-by}+c$$

Solve for $y$:

$$x-c=(a-c)e^{-by}$$

$$\frac{x-c}{a-c}=e^{-by}$$

$$\ln\left(\frac{x-c}{a-c}\right)=-by$$

$$y=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)$$

Thus, we may claim:

$$f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a-c}{x-c}\right)$$

The constant 'a' also multiplies the second term, that's why we got different results.

$$f(x)=ae^{-bx}+ac\left(1-e^{-bx}\right)$$

 

[MarkFL]

The constant 'a' also multiplies the second term, that's why we got different results.

$$f(x)=ae^{-bx}+ac\left(1-e^{-bx}\right)$$

So it does...I missed that...lemme try again:

---

We are given:

$$f(x)=a\left(e^{-bx}+c\left(1-e^{-bx}\right)\right)=a(1-c)e^{-bx}+ac$$

To find the inverse function, we can write:

$$x=a(1-c)e^{-by}+ac$$

Solve for $y$:

$$x-ac=a(1-c)e^{-by}$$

$$\frac{x-ac}{a(1-c)}=e^{-by}$$

$$\ln\left(\frac{x-ac}{a(1-c)}\right)=-by$$

$$y=\frac{1}{b}\ln\left(\frac{a(1-c)}{x-ac}\right)$$

Thus, we may claim:

$$f^{-1}(x)=\frac{1}{b}\ln\left(\frac{a(1-c)}{x-ac}\right)$$

 

[guiismiti]

So it does...I missed that...lemme try again

Done :)