Discover the Solution to e^(0.1x) = x | Precalculus Problem

[KStolen]

Sorry, I'm not sure what is considered precalculus and what isn't, so hopefully this is in the right section.

Homework Statement

e^{0.1x} = x

Homework Equations

The Attempt at a Solution

(1.105170918)^{x} = x

Unfortunately there is no context to the question, so I have no idea what approach should be taken. Can somebody explain what should be done?

 

[jackmell]

You can express the solution in terms of the Lambert-W function. Try and review this link and then study the examples in the Application section and see if you can then solve your problems.

http://en.wikipedia.org/wiki/Lambert_w_function

 

[awkward]

Sorry, I'm not sure what is considered precalculus and what isn't, so hopefully this is in the right section.

Homework Statement

e^{0.1x} = x

Homework Equations

The Attempt at a Solution

(1.105170918)^{x} = x

Unfortunately there is no context to the question, so I have no idea what approach should be taken. Can somebody explain what should be done?

There is no analytic solution in terms of familiar mathematical functions, but there is a solution in terms of the Lambert W function. See

http://en.wikipedia.org/wiki/Lambert_W_function

Transform the equation to
(-.1 x) e^{-.1 x} = -.1
then
-.1 x = W(-.1)
so
x = -10 W(-.1) \approx 1.11833

[edit]Beaten to the punch![/edit]

 

[Char. Limit]

There are two solutions, not one:

x=-10W(-.1) = 1.11833...

x=-10W_{-1}(-.1)=35.7715...

You can see that if you graph the two.

 

[jackmell]

I see no indication anywhere where only real values were requested and therefore there are actually an infinite number of solutions since the W function is infinitely-valued. For example, 44.491 - 73.0706 i is an approximation to one of the complex values. Plot the real and imaginary components of the function and you'll see what I mean. :)

But more importantly, does Stolen know how to arrive at that expression? Just divide by e^{ax} then multiply by -a and get:

-a=-axe^{-ax}

which is now in a form that the W function can be taken on both sides (see reference);

 

[KStolen]

Well I can see how to get W(-0.1) by approximating with Newton's method:

w_{n+1} = w_{n} - \frac{w_{n}e^{w_{n}}-(-0.1)}{e^{w_{n}} + w_{n}e^{w_{n}}} picking any number for w_{0}

How do I know that W_{-1}(-0.1) exists if I don't graph it, as Char.Limit says? And how do I approximate W_{-1}(-0.1)?

 

[D H]

Use a fixed point iterator. You have x=e^{0.1x}, which is of exactly the right form needed for fixed point iteration, x=f(x)[/tex]. The only concern is stability, and for that you must have |f&amp;#039;(x_0)| &amp;lt; 1 and |f&amp;#039;(x_f)| &amp;lt; 1. Here, <i>x</i>₀ is the initial guess and <i>x_f</i> is the final solution.