Transformation of a formula

[melodyw]

Non-affine function into affine function

Help. How would you transform 2 x Co^(1/2) x exp(-ktCo^(1/2))=C^(1/2) x (1+exp(-ktCo^(1/2)) into function y=ax+b, concentration C should be y, and time t should be x.

 

[Deleted member 4993]

Help. How would you transform 2 x Co^(1/2) x exp(-ktCo^(1/2))=C^(1/2) x (1+exp(-ktCo^(1/2)) into function y=ax+b, concentration C should be y, and time t should be x.

That function cannot be strictly linearized - i.e. you can linearize it if:

y = f₁(C) and

x = f₂(t)

 

[melodyw]

Think about applying "ln" to both sides.

Yes, that is the first thing i have done but i get stuck with this part (1+exp(-ktCo^(1/2)) i don't know how to get t down?

 

[Deleted member 4993]

Yes, that is the first thing i have done but i get stuck with this part (1+exp(-ktCo^(1/2)) i don't know how to get t down?

If you write: u = kt√C_o and v = √(2C_o/C)
you would get:

v = e^u + 1

Now y = v and

x = e^u

and you get

y = x + 1

 

[melodyw]

That function cannot be strictly linearized - i.e. you can linearize it if:

y = f₁(C) and

x = f₂(t)

so what should i do, how can i write it the best way, i managed to get C alone on the left side and the rest on the right side but t is still in exponent along with -k times square root from Co. What is x then? Can i write it like x=e^(t)?

 

[melodyw]

If you write: u = kt√C_o and v = √(2C_o/C)
you would get:

v = e^u + 1

Now y = v and

x = e^u

and you get

y = x + 1

Do you think this is the only way?
Our professor is quite demanding then, we can't leave it that way. We also have to make a graph out of this and the units of measurement must fit.