Solve for x, a^x = bx

[phill_roh]

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]

Can't be solved by elementary means. Perhaps related to the Lambert W. function?

 

[tkhunny]

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]

... You can pick various values for a and b and you are on your way. An obvious example might be a = b = 0. ...

0^x = (0)x = 0

x = any positive value

 

[daon]

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]

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:

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

 

[Count Iblis]

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