Solve for x, a^x = bx

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phill_roh

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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.
 
Can't be solved by elementary means. Perhaps related to the Lambert W. function?
 
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?
 
tkhunny said:
... You can pick various values for a and b and you are on your way. An obvious example might be a = b = 0. ...
Click to expand...

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

x = any positive value
 
morson said:
Can't be solved by elementary means. Perhaps related to the Lambert W. function?
Click to expand...

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.
 
Approximation methods such as Newton's method can be used.


The following problem can, however, be soved exactly.

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

\(\displaystyle \L a^{x}=b x\)

Evaluate the summation:

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

:D
 
\(\displaystyle \L\sum_{n=1}^{\infty}\frac{1}{\alpha_{n}^{2}}=b\left[b-2\log(a)\right]\) :D
 
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