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, [itex]x=f(x)[/tex]. The only concern is stability, and for that you must have $|f'(x_0)| < 1$ and $|f'(x_f)| < 1$. Here, <i>x</i>₀ is the initial guess and <i>x_f</i> is the final solution.[/itex]