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?
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6You can't. Log is not a power function. In particular, it grows more slowly than any power function, and it outputs negative values for inputs between $0$ and $1$, which power functions never do. – Nate Eldredge Dec 14 '19 at 03:53
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1@NateEldredge: I think that should be an answer. I would upvote it. I suspect the equivalence is for large $x$, so the point about negative values is not important, but OP did not define equivalent. – Ross Millikan Dec 14 '19 at 04:13
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You should say $\text{“}n$ is in $(0,\infty]\text{''}$ rather than $\text{“}n$ is $(0,\infty].\text{''} \qquad$ – Michael Hardy Dec 14 '19 at 05:06
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$\ldots$or better yet, say $\text{“for } n\in(0,\infty].\text{''}$ That avoids the use of "where" as, in effect, a quantifier. The word "where" in this context should be used for establishing notation or conventions, such as "where $a=\text{longitude},$" etc. $\qquad$ – Michael Hardy Dec 14 '19 at 05:09
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1So nitpicky $\space$ – D_S Dec 14 '19 at 05:57
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$$x^A=\ln x \implies A \ln x=\ln(\ln x) \implies A=\frac{\ln(\ln x)}{\ln x}$$ 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 $x \in (1,\infty)$.
Z Ahmed
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Note that this makes the exponent a function of $x$, so it seems a safe bet that this will not do what OP wanted it to do. – David K Dec 14 '19 at 07:02
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Yes, there is sufficient information here for OP to decide whether the idea in the question is useful, so indeed this could be the answer to the question. – David K Dec 14 '19 at 15:20
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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 $x=1$.
Nate Eldredge
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