Does there always exist a real number 'n' such that $log_{n}x$ is a whole number for any real number x?
If yes what would the function to find this number look like?
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$\begingroup$ Well, $\log_b x$ is continuous for $b \in (1, \infty)$, so given $x$, you can find $b$ such that $b^k = x$ for $k \in \mathbb{N}.$ $\endgroup$– Sean RobersonCommented Jun 15 at 5:49
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$\begingroup$ The question that you presented in your first sentence is asking for some universal value $~n~$ that will hold for any $~x.~$ The question that you presented in your second sentence, in contrast to the first sentence, is allowing the value $~n~$ to vary as $~x~$ varies. I suggest that you edit your posting to resolve the ambiguity. $\endgroup$– user2661923Commented Jun 15 at 6:44
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$\begingroup$ For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. $\endgroup$– user2661923Commented Jun 15 at 6:45
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$\begingroup$ Which question did you mean to ask: (1) Is it true that for any real number $x$, there is a real number $n$ (depending on $x$) such that $\log_n x$ is a whole number; or (2) is there a real number $n$ such that for the same value of $n$, $\log x$ is a whole number for every real number $x$? I think you mean (1) but the way the question is written, some people think it means (2). $\endgroup$– David KCommented Jun 16 at 17:50
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$\begingroup$ Actually I think you mean a third question: Given a whole number $k$ and an $x$ will there be a base $n$ so that $\log_n x = k$. the third answer to that is the $\log_n x = k\iff n^k = x$ so $n = \sqrt{k} x$. So please clarify which of the three question you mean. But every one of these questions can be answered by simply using the very definition of a logarithm. $\log_n x = k \iff n^k=x$. $\endgroup$– fleabloodCommented Jun 16 at 22:48
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2 Answers
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Yes indeed there is.
if you take $n=x$
$log_x(x) = 1$.
You can also just take $n$ to be any root of $x$
$log_\sqrt{x}(x)=2$
$log_\sqrt[3]{x}(x)=3$
and so on.
This of course only works for positive real numbers
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$\begingroup$ I think OP asked a constant n value not a function of x. I don't think it's possible. But his next sentence changes his enquiry. You're correct in that case :) $\endgroup$– GwenCommented Jun 15 at 12:33
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$\begingroup$ An assertion of the form "there exists $n$ such that ... for every $x$" is ambiguous. It could mean: "there exists $n$ such that for every $x$ ..." OR it could mean "for every $x$ there exists $n$ such that ...". $\endgroup$– GEdgarCommented Jun 16 at 16:31
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$\begingroup$ A constant value $n$ where $\log_n x \in \mathbb N$ for all $x \in \mathbb R$ is impossible as $\log_n n^w = w$. and we just have to take $w\not \in \mathbb N$. $\endgroup$ Commented Jun 16 at 17:07
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What exactly are you asking?
Are you asking: If you are given a number $n$ can you always find an $x$ so that $\log_n x$ is a natural number?
If so the answer is yes. To solve $\ln x = 3$ we just do $e^3 = x$ so $x=\sqrt[3] e$ and for any $\log_n x = k\in \mathbb N$ then $n^k =x$ and $x = \sqrt[k]n$.
Or are you asking. Is there a $n$ so that for every $x\in \mathbb R$ then $\log_w x$ is a whole number?
If so the answer to that is not by a long shot.
Consider any number $n$ so that $n\ne 1$. Then consider where $v$ is not a whole number. Let $x = n^v$. Then $\log_n x = \log_n n^v = v$. Which... is not a whole number.
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This all really dances around the idea of whether $\log_n x$ and $n^x$ are actually well defined and if so how (they are) and I avoided the issue of "Gee, if $1^x = 1$ for all $x$ can we say $log_1 1$ can be any value" (you can't). Those are other issues.
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$\begingroup$ If you feel the need to begin your answer by writing several questions which are meant to clarify what the question is, please do not post an answer to the question, and instead use the comments to ask for clarification from the asker. $\endgroup$– Xander Henderson ♦Commented Jun 16 at 19:21