# The value for t

Tags:
1. Oct 15, 2016

### Esas Shakeel

• Member warned that an effort must be shown
1. The problem statement, all variables and given/known data

s = 0.5(e^t - t - 1)
t=?
when s=18
2. Relevant equations
the use of natural log probably

3. The attempt at a solution
I tried using the natural log rule but im getting stuck. Please help me out here. If someone could give a step by step solution,I'll be rreally glad, Cheers

2. Oct 15, 2016

### Math_QED

It's a hard forum requirement that you show some work, what you tried, what you thought, etc. Nobody here will write out the entire solution for you.

3. Oct 15, 2016

### Esas Shakeel

@Math_QED

18=0.5e^t - t -1

19 = 0.5e^t -t

ln(19) = tln(0.5)-lnt ??

4. Oct 15, 2016

### Krylov

Are you looking for only positive solutions for $t$? How many solutions do you think there are in total? Make a graph.
It is probably not possible to find an expression for $t$ in terms of standard functions, so you may have to call upon a numerical root finder.

5. Oct 15, 2016

### Esas Shakeel

Here, t represents time so if theres a easier way to find only the positive values, then i'd like to know that please :(

6. Oct 15, 2016

### Krylov

There is no solution (positive or otherwise) in terms of elementary functions, so if this is a homework problem that asks you to find an exact solution, you may have to re-read the problem.
If this is indeed the correct equality, then you could proceed to make a graph to get an idea of where the solution may be. Then you can invoke a root finding to determine it approximately.

7. Oct 15, 2016

### Esas Shakeel

@Krylov this is the solution. but im not sure how he got to the value of t :/

8. Oct 15, 2016

### Esas Shakeel

Alright sorry i didnt put the bracket. My apologies :p

9. Oct 15, 2016

### Krylov

No, $t = 3.7064$ does not satisfy $18 = 0.5(e^t - t - 1)$. (Try it.) It satisfies this equality only approximately and it was probably found using a numerical root finder or by reading it off from a sufficiently detailed graph.

10. Oct 15, 2016

### Esas Shakeel

What the heck?! I've wasted pages just to get the answer but i couldnt. I'll sue the author for this! lol . I'll take this question to my dynamics teacher and see what he comes up with. Thanks man! @Krylov

11. Oct 15, 2016

### pasmith

You can't solve $$e^t - t = 37$$ analytically; if you have a calculator, then the iteration $$t_{n+1} = \ln(t_n + 37)$$ with $t_1 = 0$ converges rapidly to $t = 3.706384959$ to the precision my calculator will display. (How did I know to try that instead of $t_{n+1} = e^{t_n} - 37$? For the iteration $t_{n+1} = f(t_n)$ to converge to a solution of $t = f(t)$ which you know to exist then you must have $|f'(t)| < 1$ near that solution. $\ln(t + 37)$ has derivative $0 < 1/(t + 37) < 1$ for all $t > 0$; $e^t - 37$ has derivative $e^t > 1$ for all $t > 0$.)

12. Oct 15, 2016

### Ray Vickson

No, absolutely NOT.
$$\ln(0.5 e^t - t) \neq t \ln(0.5) - \ln(t)$$.
Never, ever write $$\ln(a \pm b) = \ln(a) \pm \ln(b) \; \Longleftarrow \; \text{FALSE!}$$
The correct property of logarithms is
$$\ln(a \times b) = \ln(a) + \ln(b)\; \Longleftarrow \; \text{True!}$$
Anyway, this problem has no solution in elementary functions, but it can be solved in terms of the so-called Lambert W-function. That is a decidedly non-elementary function, whose evaluation will not appear on any calculator button and probably not even in any spreadsheet.

It is probably easiest to just look for a numerical solution using one of the many numerical root-finding methods available.

13. Oct 15, 2016

### Esas Shakeel

But i just used the natural log power rule
ln(x^y) = y ∙ ln(x)
how can that be wrong?
@Ray Vickson

14. Oct 15, 2016

### Ray Vickson

No, you did not; I was responding to your post where you wrote

19 = 0.5e^t -t

ln(19) = tln(0.5)-lnt ??

That was a "cut and paste" of your actual writing.

Basically, you were saying that $\ln(.5 e^t -t) = \ln(0.5 e^t) - \ln(t)$, which is definitely wrong.

15. Oct 15, 2016

### Staff: Mentor

Noted.

16. Oct 16, 2016

### Esas Shakeel

@Ray Vickson
Oh i get it! Yeah you're right. Natural log really confuses me even though im well aware of the existence of these simple rules :p