What can you raise $x$ to in order for it to be equivalent to $\ln(x)$?

Question

Take, for example, $\frac{\ln(x)}{x^5}$. Can this be simplified into $x^n$?

If there exists an answer, can it be generified to work for $\log_n x$ where $n$ is $(0, \infty]$?

If it can't be, why not?

Answer 1

$$x^A=\ln x \implies A \ln x=\ln(\ln x) \implies A=\frac{\ln(\ln x)}{\ln x}$$ Yes, it can be generalized to $\log_n(x)$, where the base $n \in (0, 1) \cup (1,\infty)$. Further, if $n \in (0,1)$, then $x$ should also in in $(0,1)$. Similarly, if $n \in (1, \infty)$ so should be $x \in (1,\infty)$.

Answer 2

If you mean for $n$ to be a constant, then this is impossible. There is no constant value of $n$ such that $\frac{\ln x}{x^5} = x^n$ for every $x$. The logarithm function simply isn't a power function.

You can get a pretty good sense of this if you graph them both for various values of $n$. You'll see that they are never even close to being the same function.

Algebraically, maybe the simplest way to prove this is to just see what happens when $x=1$. Then $\frac{\ln 1}{1^5} = 0$, while $1^n = 1$ no matter what $n$ is.

If $n$ is allowed to depend on $x$, then you can do it for all $x \ne 1$ as in Zafar Ahmed's answer. However, as Zafar notes it still doesn't work for $x=1$.