Is x=1 the Solution to nx ~ n ln(n) as n Approaches Infinity?

[japplepie]

How do I solve for x in the relationship below:

n^x ~ n ln(n), as n -> infinity

The answer that I'm getting is x=1, but that must be wrong since 1 ~ ln(n) as n-> infinity is wrong.

 

[pasmith]

How do I solve for x in the relationship below:

n^x ~ n ln(n), as n -> infinity

The answer that I'm getting is x=1, but that must be wrong since 1 ~ ln(n) as n-> infinity is wrong.

There is no solution: \frac{\ln n}{n^k} \to \begin{cases} 0 & k > 0 \\ \infty & k \leq 0 \end{cases}

 

[japplepie]

oh boy