Is There a Trick to Rewrite This Equation in Terms of t?

[SUDOnym]

Hi

I want to rewrite the following equation in terms of t:

X=\exp(-At)-\exp(-Bt)

Where X,A and B are positive numbers.

The problem is, I have a -exp() term... so I can't take the natural log of this... ie. as far as I am aware the following is not allowed:

\ln(-\exp(-Bt))

So is there any trick I can use in order to rewrite this equation in terms of t?

 

[SUDOnym]

of course if I wasn't such an idiot I would've seen that I simple have to add exp(-Bt) to both sides - I retract this thread!

 

[SUDOnym]

apologies! - I in fact still do not know how to solve it, although I thought I did for a minute..:
if I do what I said in my second post, it leads to (EQN1):

X+\exp(-Bt)=\exp(-At)

and taking the natural log then goes to (EQN2):

\ln(X+\exp(-Bt))=-At

***please also note a correction to my first post, X is a negative real number - I said it was positive in my first post. And also recall X, A and B are known numbers (A and B are positive).

finally note that I also know that the quantity on the LHS of (EQN1) is greater than 0, ie:

X+\exp(-Bt)>0

so to summarise: I don't know how to simplify the expression on the LHS of (EQN2)..

 

[Dickfore]

x = e^(-at) - e^(-bt)

take the natural log of both sides; it is -ln(e^(-bt)) which makes it bt :)

ln(x)=-at+bt

This is wrong! If you exponentiate back, the r.h.s. becomes:

<br /> \exp{(-a t + b t)} = \exp{(-a t)} \exp{(b t)} \neq \exp{(-a t)} - \exp{(-b t)}<br />

The point is that this equation is not solvable in closed form using elementary functions.

 

[t.francis]

^ correct. my bad. It is 2 int morning! :P

 

[SUDOnym]

Thanks to both repliers: Dickfore and t.francis...
but there seems to be a glitch with the thread - I am only seeing one reply from both Dickfore and t.francis but what they have said in these posts suggests that they left earlier posts as well... ie. the post from Dickfore reads:

Originally Posted by t.francis View Post

x = e^(-at) - e^(-bt)

take the natural log of both sides; it is -ln(e^(-bt)) which makes it bt :)

ln(x)=-at+bt

This is wrong! If you exponentiate back, the r.h.s. becomes:

exp(−at+bt)=exp(−at)exp(bt)≠exp(−at)−exp(−bt)


The point is that this equation is not solvable in closed form using elementary functions.

followed by a post from t.francis:

^ correct. my bad. It is 2 int morning! :P

So presumably you both left earlier posts that might be helpful?

Finally @Dickfore

The point is that this equation is not solvable in closed form using elementary functions.

Can you elaborate on what that means exactly - what is "closed form"? and are saying that I can only solve this equation numerically?

 

[uart]

Can you elaborate on what that means exactly - what is "closed form"? and are saying that I can only solve this equation numerically?

In general yes, you'd have solve numerically.

There a few special cases, where there is a simple relationship between "A" and "B", in which you could make it into a quadratic or cubic and solve in closed form. Eg A=-B or A=2B or B=2A let's you write it as a quadratic while A=3B or B=3A or A=-2B or B=-2A let's you write it as a cubic.

 

[Dickfore]

Can you elaborate on what that means exactly - what is "closed form"? and are saying that I can only solve this equation numerically?

Closed form means you cannot express the solution as:
<br /> t = f(X; A, B)<br />
where f(.; . , . ) is an "[URL function
[/URL]