Exponential equation: solve y = a (1 - e^{-bx}) for b

[SHS]

Hello,

I have an exponential equation with 2 unknowns (a and b). Also, I have 2 points (x1 , y1) and (x2 , y2) to put in the equation. But, I can not solve the equation with respect my points:

y = a (1 - e^-bx)

In fact, I need to solve the equation in a way that I'd have 2 equations for a and b:

a = .....
b = .....

Does anyone know how to solve it?

 

[Deleted member 4993]

Hello,

I have an exponential equation with 2 unknowns (a and b). Also, I have 2 points (x1 , y1) and (x2 , y2) to put in the equation. But, I can not solve the equation with respect my points:

y = a (1 - e^-bx)

In fact, I need to solve the equation in a way that I'd have 2 equations for a and b:

a = .....
b = .....

Does anyone know how to solve it?

y₁/y₂ = [1- e^(-b*x₁)]/[1- e^(-b*x₂)]

Now you have one equation and one unknown (b) ... and continue....

 

[stapel]

I have an exponential equation with 2 unknowns (a and b). Also, I have 2 points (x1 , y1) and (x2 , y2) to put in the equation. But, I can not solve the equation with respect my points:

y = a (1 - e^-bx)

In fact, I need to solve the equation in a way that I'd have 2 equations for a and b:

a = .....
b = .....

Does anyone know how to solve it?

When you say that you "have two points", do you mean that you have specific x- and y-values? If so, then plug these values into the generic equation. This will give you two equations in two unknowns, being "a" and "b". Solve the system to get the specific form of your equation.

 

[SHS]

y₁/y₂ = [1- e^(-b*x₁)]/[1- e^(-b*x₂)]

Now you have one equation and one unknown (b) ... and continue....

You're right. But, this is the way that I've already gone and I couldn't solve the rest.I mean I couldn't separate "b".

 

[stapel]

You're right. But, this is the way that I've already gone and I couldn't solve the rest.I mean I couldn't separate "b".

If you'd like, you can post your work so far, showing all of your steps, and we'll be glad to try to help you get un-stuck.

 

[SHS]

If you'd like, you can post your work so far, showing all of your steps, and we'll be glad to try to help you get un-stuck.

y = a (1 – e^-bx)
we have (x1, y1) and (x2, y2), as two points of the equation:
y₁ = a (1 – e^-bx1) & y₂ = a (1 – e^-bx2)
Thus,
y₁/y₂ = (1 – e^-bx1) / (1 – e^-bx2)
y₁ (1 – e^-bx2) = y₂ (1 – e^-bx1)
y₁ - y₁ (e^-bx2) = y₂ - y₂ (e^-bx1)
Now, I need to have an equation based on y = a (1 – e^-bx)
we have (x1, y1) and (x2, y2), as two points of the equation:
y₁ = a (1 – e^-bx1) & y₂ = a (1 – e^-bx2)
Thus,
y₁/y₂ = (1 – e^-bx1) / (1 – e^-bx2)
y₁ (1 – e^-bx2) = y₂ (1 – e^-bx1)
y₁ - y₁ (e^-bx2) = y₂ - y₂ (e^-bx1)
Now, I need to have an equation based on x₁, y₁, x₂ and y₂ , equals to b:
b = ……… ?
, which is equal to b:
b = ……… ?

Of course, I don't want to put numbers instead of x₁, y₁, x₂ and y₂, right now. I just want to have the equation of "b" to write a C code with that.

 

[stapel]

y = a (1 – e^-bx)
we have (x1, y1) and (x2, y2), as two points of the equation:
y₁ = a (1 – e^-bx1) & y₂ = a (1 – e^-bx2)
Thus,
y₁/y₂ = (1 – e^-bx1) / (1 – e^-bx2)
y₁ (1 – e^-bx2) = y₂ (1 – e^-bx1)
y₁ - y₁ (e^-bx2) = y₂ - y₂ (e^-bx1)...

That's kinda nasty. I'd assumed you'd have values for the points (y₁, x ₁ ) an d (y ₂ , x₂ ). Lacking that, I'd go like this:

. . . . \(\displaystyle y_1\, -\, y_1\, e^{-bx_2}\, =\, y_2\, -\, y_2\, e^{-bx_1}\)

. . . . \(\displaystyle y_1\, -\, y_2\, =\, y_1\, e^{-bx_2}\, -\, y_2\, e^{-bx_1}\)

Then use the fact that:

. . . . \(\displaystyle a^{mn} = (a^m)(a^n)\)

...to get:

. . . . \(\displaystyle y_1\, -\, y_2\, =\, y_1\, e^{b}\, e^{-x_2}\, -\, y_2\, e^{b}\, e^{-x_1}\)

. . . . \(\displaystyle \dfrac{y_1\, -\, y_2}{e^{b}}\, =\, \dfrac{e^{b}\, (y_1\, e^{-x_2}\, -\, y_2\, e^{-x_1})}{e^{b}}\)

. . . . \(\displaystyle \dfrac{y_1\, -\, y_2}{y_1\, e^{-x_2}\, -\, y_2\, e^{-x_1}}\, =\, e^{b}\)

Then log either side to get what you're seeking.

 

[SHS]

That's kinda nasty. I'd assumed you'd have values for the points (y₁, x ₁ ) an d (y ₂ , x₂ ). Lacking that, I'd go like this:

. . . . \(\displaystyle y_1\, -\, y_1\, e^{-bx_2}\, =\, y_2\, -\, y_2\, e^{-bx_1}\)

. . . . \(\displaystyle y_1\, -\, y_2\, =\, y_1\, e^{-bx_2}\, -\, y_2\, e^{-bx_1}\)

Then use the fact that:

. . . . \(\displaystyle a^{mn} = (a^m)(a^n)\)

...to get:

. . . . \(\displaystyle y_1\, -\, y_2\, =\, y_1\, e^{b}\, e^{-x_2}\, -\, y_2\, e^{b}\, e^{-x_1}\)

. . . . \(\displaystyle \dfrac{y_1\, -\, y_2}{e^{b}}\, =\, \dfrac{e^{b}\, (y_1\, e^{-x_2}\, -\, y_2\, e^{-x_1})}{e^{b}}\)

. . . . \(\displaystyle \dfrac{y_1\, -\, y_2}{y_1\, e^{-x_2}\, -\, y_2\, e^{-x_1}}\, =\, e^{b}\)

Then log either side to get what you're seeking.

OOOMMMMGGGG!!!

I didn't use \(\displaystyle a^{mn} = (a^m)(a^n)\) !!! Why?!!!

Thank you very much Kinda...