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phill_roh
05-13-2007, 01:52 AM
My friend needs help answering this question for his math class. Don't know why it's so important, but he said he'll buy me a bucket of ice cream if I can get the answer for him :-). I promise to share with anybody who helps me! He needs the method to solve the following problem:

Solve for x

a^x = bx.

morson
05-13-2007, 08:12 AM
Can't be solved by elementary means. Perhaps related to the Lambert W. function?

tkhunny
05-14-2007, 12:08 PM
Can't generally be solved by elementary means. You can pick various values for a and b and you are on your way. An obvious example might be a = b = 0. Can you think of others?

skeeter
05-14-2007, 01:11 PM
... You can pick various values for a and b and you are on your way. An obvious example might be a = b = 0. ...

0<sup>x</sup> = (0)x = 0

x = any positive value

daon
05-14-2007, 01:28 PM
Can't be solved by elementary means. Perhaps related to the Lambert W. function?

I looked at Lambert's function and can't find an application to this problem. Unless there is an extension of this function dealing with arbitrary logarithms.

Count Iblis
05-14-2007, 06:33 PM
Approximation methods such as Newton's method can be used.

The following problem can, however, be soved exactly.

Suppose \L\alpha_{n} for \L n=1\ldots\infty are all the complex solutions of the equation:

\L a^{x}=b x

Evaluate the summation:

\L\sum_{n=1}^{\infty} \frac{1}{\alpha_{n}^{2}}

:D

Count Iblis
05-15-2007, 09:11 AM
\L\sum_{n=1}^{\infty}\frac{1}{\alpha_{n}^{2}}=b\le ft[b-2\log(a)\right] :D