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I would like to know if there is a name given to equations like the following:

$$-31=\frac{-39.2}{x}(1-e^{\frac{x}{4}})$$

or more generally:

$$a=\frac{b}{x}c^x$$

The formatting doesn't really matter, I'm just talking about an equation where there are variables both in an exponent and in another term in a base. Another example would be this:

$$20x^2 = e^x$$

I've searched for things like "composite exponential functions" but to no avail.

Also, I guess I could solve my example by rearranging so that there is an expression containing x on the left and an exponential one on the right. Then I could graph each side and see where they equal. However, is there a way to do it analytically?

Perhaps I'm not explaining this as clearly as I could be, but I would appreciate any advice you could offer.

Thanks!

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Welcome to the world of Lambert function !

If you consider equation $$a=\frac{b \, e^{c x}}{x^d}$$ the solution is $$x=-\frac{d}{c} W\left(-\frac{c }{d}\left(\frac{b}{a}\right)^{1/d}\right)$$ where $W(z)$, the Lambert function, is such that $z=W(z)\,e^{W(z)}$.

The Wikipedia page will provide you many examples of the series of manipulations to be done.

In the real domain, $W(z)$ exists if $z \geq -\frac 1 e$ and, for $z<0$, it shows two branches.

Considering the example you give $a=20$, $b=1$, $c=1$, $d=2$, the result will be $$x=-2 W\left(-\frac{1}{4 \sqrt{5}}\right)$$ The Wikipedia page will also provide you series expansions for computing the value of $W(z)$. Using for example $$W(z)=z-z^2+\frac{3 z^3}{2}-\frac{8 z^4}{3}+\frac{125 z^5}{24}+O\left(z^6\right)$$ amking for your example $$x=\frac{7936+31321 \sqrt{5}}{307200}\approx 0.253815$$ while the exact value should be $\approx 0.253871$.

In fact, sooner or later, you will learn that any equation which can write $A+Bx+C\log(D+Ex)=0$ has solution(s) in terms of Lmabert function.

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Thanks for the detailed answer!

I was a bit confused at first because the Wikipedia page mentioned that it is defined on the complex plane. However, I reread your comment and saw what you said about it being defined for real numbers with domain restrictions, and things are making more sense now.

Hopefully I will get to study Lambert functions in more depth at some point in the future, but, for now, I'm glad that I understand the basic concept.

I think I would still be searching around for an answer had I just continued trying to Google things, so much appreciation for helping out!

-I

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  • $\begingroup$ I think it would have been better if this was a (or a couple of) comment(s) rather than an answer. Try upvoting (pressing the up button) and accepting (the tick mark)the answer if Claude's answer helped you. $\endgroup$ – Mohammad Zuhair Khan Jan 21 '18 at 5:48

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