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To find how many $n$ are such that $\log_{n}{3^{2013}}$ is an integer and $n$ is a natural number.

So somehow I need to make sure $\log_{n}{3}$ is an integer for natural values of $n$

Other than $3$, I can't think of any other value of $n$ that satisfies this. But $8$ such values are possible. How can that be ??

Any help is appreciated.

Blue
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Glowingbluejuicebox
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    Hint: take a look at the prime factorization of 2013 – Vectorizer Apr 15 '21 at 08:11
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    $\log_{n}{3}$ doesn't have to be an integer, for example take n = $3^{2013}$. – lc2r43 Apr 15 '21 at 08:22
  • Well $\log_{3^{2013}} 3^{2013} = 1\in \mathbb N$. – fleablood Apr 15 '21 at 08:34
  • @fleablood I think what lc2r43 meant is: It is sufficient but not necessary – Vectorizer Apr 15 '21 at 12:34
  • @Vectorizer I wasn't responding to lc2r43. I was making the same point lc2r43 was making but I hadn't read his comment. I was reponding to the OP's comment of not being able to think of any number but $n$. I was pointing out ... well, if $n = 3^{2013}$ then $\log_n 3^{2013} = 1$. I hadn't scrutinized that that was the exact example lc2r43 had given ot $\log_n 3$ needing be an integer. ... That was a bit bad of me... but ... I think it still hits not a pail of a cold water to the face that $\log_n 3^{2013} = 1$ if $n = 3^{2013}$ certainly gives an alternative to $n=3$. – fleablood Apr 15 '21 at 17:50

1 Answers1

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$\log_n 3^{2013} = k\iff n^k = 3^{2013}$.

So if $n$ is natural it's only prime factor can be $3$. So $n = 3^m$ for some $m$ so $(3^m)^k = 3^{2013}$.

In other words $mk = 2013$. And $m$ can be any factor of $2013$.

The prime factorization of $2013 = 3 \cdot 11\cdot 61$ and there are $8$ factors: $1,3, 11, 61, 33, 183, 671, 2013$.

So we can have $n = 3^{\text{any of those powers}}$.

For example $\log_{3^{2013}} 3^{2013} = 1$.

And $\log_{3^{61}} 3^{2013} = 33$ (Because $(3^{61})^{33} = 3^{2013}$.

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Alternatively if

$\log_n 3^{2013} =k$ then

$2013 \log_n 3 = k$

$\log_n 3 =\frac k{2013}$

Now there is a log identity that $\log_a b = \frac 1{\log_b a}$ (use the change of base rule: $\log_a b = \frac {\log_b b}{\log_b a}= \frac 1{\log_b a}$.

So $\frac {2013}{k} = \log_3 n$.

So $n = 3^{\frac {2013}k}$.

If $ \frac {2013}k$ is a non-integer rational then $ 3^{\frac {2013}k}$ can not be an integer.

So if $n$ is natural then $3^{\frac {2013}k}$ where $k$ is an integer that divides evenly into $2013$. As there are $8$ such integer factors there are $8$ such numbers.

fleablood
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