Find the Value of t Using Natural Log for s=18 in s = 0.5(e^t - t - 1)"

[Esas Shakeel]

Homework Statement

s = 0.5(e^t - t - 1)
t=?
when s=18

Homework Equations

the use of natural log probably

The Attempt at a Solution

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

 

[member 587159]

Homework Statement

s = 0.5(e^t - t - 1)
t=?
when s=18

Homework Equations

the use of natural log probably

The Attempt at a Solution

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

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.

 

[Esas Shakeel]

@Math_QED

18=0.5e^t - t -1

19 = 0.5e^t -t

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

 

[S.G. Janssens]

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.

 

[Esas Shakeel]

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

 

[S.G. Janssens]

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.

 

[Esas Shakeel]

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

 

[Esas Shakeel]

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

 

[S.G. Janssens]

View attachment 107467@Krylov this is the solution. but I am not sure how he got to the value of t :/

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.

 

[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

 

[pasmith]

You can't solve <br /> e^t - t = 37 analytically; if you have a calculator, then the iteration <br /> 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 <br /> 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&#039;(t)| &lt; 1 near that solution. \ln(t + 37) has derivative 0 &lt; 1/(t + 37) &lt; 1 for all t &gt; 0; e^t - 37 has derivative e^t &gt; 1 for all t &gt; 0.)

 

[Ray Vickson]

@Math_QED

18=0.5e^t - t -1

19 = 0.5e^t -t

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

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.

 

[Esas Shakeel]

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

 

[Ray Vickson]

View attachment 107467@Krylov this is the solution. but I am not sure how he got to the value of t :/

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

 

[Mark44]

It's a hard forum requirement that you show some work, what you tried, what you thought, etc.

Noted.

 

[Esas Shakeel]

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