How can we algebraically find the intersection of an exponential function y=a^x with the straight line y=x? In other words, what is the solution to the equation a^x = x ?

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- Mar 1st 2008, 01:11 AM #1

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- Mar 1st 2008, 01:42 AM #2

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First there is no guarantee that there are any real solutions to $\displaystyle a^x=x$. In fact if $\displaystyle \ln(a)>1$ it is obvious that there are no real solutions.

If $\displaystyle a<1$, then put $\displaystyle b=1/a$ and the equation becomes:

$\displaystyle e^{-\ln(b)x}=x$

which we can rearrange to:

$\displaystyle [\ln(b)~x]~e^{\ln(b)~x}=\ln(b)$

and this has solution:

$\displaystyle x=\frac{W(\ln(b))}{\ln(b)}$

or equivalently:

$\displaystyle x=-~\frac{W(-\ln(a))}{\ln(a)}$

where $\displaystyle W$ is Lambert's W function.

(If we are not concerned with real solutions then the above works period, though you have to keep an eye on the multiple values of the W function)

RonL

- Mar 1st 2008, 02:13 AM #3

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To demonstrate this working see this:

Code:This is EULER, Version 2.3 RL-06. Type help(Return) for help. Enter command: (16777216 Bytes free.) Processing configuration file. Done. >load "C:\Program Files\EulerRL\Euler Utils\LambertW.e"; > >a=0.1; >x=-LambertW(-log(a))/log(a) 0.399013 > > > >a^x-x 5.55112e-017 >

- Mar 1st 2008, 03:26 AM #4
Some replies do not deserve an unthanked fate. For my money, the above two are in that class. Hence the pre-emptive safe-guard.

(Would've added rep but apparently I've got to spread a bit more first).

And this thread is certainly as good as any to direct the interested reader to the following references:

http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf

New analytic solution to classic problem?

- Mar 1st 2008, 11:37 PM #5

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