# Dealing with natural logs

1. Oct 8, 2011

### 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?

2. Oct 8, 2011

### 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!

3. Oct 8, 2011

### 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)..

4. Oct 8, 2011

### Dickfore

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

$$\exp{(-a t + b t)} = \exp{(-a t)} \exp{(b t)} \neq \exp{(-a t)} - \exp{(-b t)}$$

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

5. Oct 8, 2011

### t.francis

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

6. Oct 9, 2011

### 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:

followed by a post from t.francis:

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

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

7. Oct 9, 2011

### uart

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 lets you write it as a quadratic while A=3B or B=3A or A=-2B or B=-2A lets you write it as a cubic.

8. Oct 9, 2011

### Dickfore

Closed form means you cannot express the solution as:
$$t = f(X; A, B)$$
where $f(.; . , . )$ is an "[URL [Broken] function
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Last edited by a moderator: May 5, 2017