How do I solve an equation involving $e^t$ and $t$ on one side? $18=0.5e^t-0.5t-0.5$

The equation is:

$$18=0.5e^t-0.5t-0.5$$

How do I solve for $t$? The answer manual gives $3.706$, but it gives no explanation on how to it got there.

This is from a recommended dynamics problem, while preparing for a test which does not allow any more than simple scientific calculators. It can be solved graphically, but that wouldn't be available during an exam.

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Answer is $\approx 3.451525244$. –  Shahar Apr 5 at 21:40
guys, sincere apologies, it was a minus and not plus..... –  MikeFoxtrot Apr 5 at 21:41
@Amzoti think you posted an answer for a few moments so i got an idea. we most probably won't have time for it, they have more junk to throw at us instead :) –  MikeFoxtrot Apr 5 at 22:10
No (dynamical-systems) here. –  Did Apr 17 at 7:29
@Did sorry, was from a dynamics problem, it was the most fitting tag i could find. I'm not sure what it should go under. –  MikeFoxtrot Apr 17 at 11:15

Newtons method generally works but its also kind of boring. Since you are in a dynamics class you should be familiar with the concept of a fixed point. In this case we will $$invent$"$ a sequence for which the solution of this equation is a fixed point.

Define $y(t) = 0.5e^t-0.5t-0.5$ and let $t_c$ be a number such that $y(t_c)=18$. Some simple estimations,

$$y(1) = 0.5 e^{1} -0.5(1)-0.5 \approx 0.5(3)-1 = 0.5$$ $$y(3)= 0.5e^3 -0.5(3)-0.5 \approx 0.5(27)-2=16.5$$

So we expect the intersection to be just past $t=3$, this is what we will use as our seed value later. I used these values to make a very rough sketch of the graph which can be seen below.

Now we will make a sequence which converges to the solution. One way is to make the recurrence shown below where $a(t_c)=0$ when the solution is reached. Notice that it is important that a(t) be positive when $t<t_c$ and negative when $t>t_c$ otherwise the fixed point won't be an attractor.

$$t_{n+1} = t_n + a(t_n)$$

There are many choices for $a(t)$. A few options are listed below.

• $a(t)=18-y(t)$
• $a(t) = \tan^{-1}(18-y(t))$
• $a(t) = 1-y(t)/18$

Another possible sequence to try is one which multiplies the old values of $t$. Looking at the sequence below we can see that $b(t_c)=1$. It should be clear that for $t_c$ to be an attractor we must have that $b(t)$ is greater than 1 when $t$ is less than $t_c$, and less than $1$ when $t$ is greater than $t_c$.

$$t_{n+1} = t_n b(t_n)$$

Some possible choices here are ,

• $b(t) = 18/y(t)$
• $b(t) = \frac{4}{\pi} \tan^{-1}(18/y(t))$
• $b(t) = \frac{1}{2-18/y(t)}$

Try playing around with these and see if you can invent your own.

Implementing the first example for $a(t)=18-y(t)$ one finds that with a seed value of $t=3$ the sequence quickly diverges. When you run into this you need to find some way of keeping $a(t)$ from being too large near the intersection. One way is to use the inverse tangent function. In this case it is as easy as dividing our choice by 18.

Using $t_0=3$ and $a(t)=1-y(t)/18$ I get,

• $t_0=3$
• $t_1=3.553\dots$
• $t_2=3.7095 \dots$
• $t_3=3.70605\dots$

• $\vdots$

• $t_{10}=3.706384959$

Which is the limit of the precision of my calculator.

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Unfortunately, this is a nasty result that uses the Lambert-W Function.

Here is a Wolfram Alpha link for the analytic result. Click on approximate form of the root of $t \approx 3.7063849587374473209$.

See this paper for ways to approximate this function.

Normally, you could use some root finding algorithm like Newton's, Secant, Bisection, or many others.

You can pretty quickly and easily do Newton's Method (although it does not work on this nasty problem), or maybe even program that algorithm into the calculator you have (and depending on model, may already have such algorithms).

However, my guess would be that the problem uses something that can be factored and you can find the roots quickly or just plot it and eyeball the result as best you can for an approximate root and refine that numerically.

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$18=0.5e^t-0.5t-0.5$ can be rewritten as $t = e^t-37$ or $t = ln(t+37)$

Both $f(t)=e^t-37$ and $g(t)=ln(t+37)$ are increasing, and one is the inverse of the other, so one will cross $h(t)=t$ from above and one from below. A little consideration will tell you it's $g(t)$ that crosses from above.

This tells us if we iterate the function $g(t) = ln(t+37)$ we will converge on the answer.

If we guess $3$ we get the iterates:

$3.00000000000000\\ 3.68887945411394\\ 3.70595482302821\\ 3.70637439189438\\ 3.70638469915054\\ 3.70638495236039\\ 3.70638495858079\\ 3.70638495873360\\ 3.70638495873735\\ 3.70638495873744\\ 3.70638495873745\\ 3.70638495873745\\$

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