Re: Solving for constants

If A, B, and C are constants then AB is a constant (let's call it 'D') and also BC is a constant (call it 'F'). So the equation is better written as $\displaystyle x(t) = De^{Ft}$. You can find the value for D using the value of x(t) at t=0, since x(0) = D. Then if there is at least one other point on the graph where for a known value of 't' you know the value for x(t), you can determine the value for F using the relationship $\displaystyle F = \frac {\ln(x(t))-ln(D)}{t} $.

Re: Solving for constants

Hi,

Thanks. I've come to a hurdle as my equation is a bit more complex:

$\displaystyle x(t) = [(A + B)/AB] * exp[At/C]$

What I have done so far is:

Let $\displaystyle K_{1} = [(A + B)/AB], K_{2}=A/C$, then using a graph of the function I get $\displaystyle K_{1} = 0.2, K_{2} = 1e-5$. I can't see where to go from here - where do I get a third equation from?

Thanks :)

Re: Solving for constants

You still have too many constants. You could set 'A' to whatever value you want, and then determine values for B and C from:

$\displaystyle B= \frac A {K_1 A-1}$

$\displaystyle C = \frac A {K_2}$

For example if you set A=1, then using your values for K1 amd K2 you get B = -1.25 and C = 1/(e-5).

Re: Solving for constants

Thanks.

The question asks me for the specific value of each of the three constants, so I can't set any constant to an arbitrary value.

I must have gone wrong somewhere in the working...will take a look.

Re: Solving for constants

What you need to look for is another relationship (i.e. equation) involving at least 2 of A, B, and C. The graph can *only* produce K1 and K2 for you, and you've already used it to determine them, so the graph will be of no further use to you.

Since you need it in exactly that form, with A, B, and C, those variables are meant to represent something. The additional relationship you seek (assuming the problem is properly defined), if not some explicit condition you've yet to mention, will likely come somehow from the meaning of the A, B, and C. The graph has done all it can for you.