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I wonder if it's possible to approximate $\log(n)$, n integer, by using a linear combination of $\log(2)$ and $\log(3)$.

More formally, given integer $n$ and and real $\epsilon>0$, is it always possible to find integer $x,a,b$ where:

$\left|n^x-2^a 3^b\right|<\epsilon$

For example, I can approximate $11$ by $2^{-33} 3^{23}=10.959708460955880582332611083984375 \approx 10.96.$

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    $a$ and $b$ could be negative integers, that can make differences less than 1.2012-06-18

1 Answers 1

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Yes.

Let $a=\frac{\log(2)}{\log(3)}$. Then, $a$ is irrational, thus by Dirichclet Theorem, the set $\{ ma+n | m,n \in Z \}$ is dense. Thus, there exists some $m,n \in Z$ so that

$\left| \frac{\log(n)}{\log(3)} - ma -k \right| < \frac{\epsilon}{\log(3)}$

Multiply by $\log(3)$ and you are done.

P.S. It is irrelevant that $n$ is integer. Also, the proof works if you replace $2$ and $3$ by any numbers $x,y$ so that $\log_x(y)$ is irrational.

P.P.S. I think that for $n$ positive integer, it is enough to use one $\log(2)$. Indeed, if $n$ is a power of 2, you are done, otherwise, $\frac{\log(n)}{\log(2)}$ is irrational, and then the set $m\frac{\log(n)}{\log(2)} - k$ is dense. Thus, you can find some integers so that

$\left|m\frac{\log(n)}{\log(2)} - k \right| < \frac{\epsilon}{\log(2)}$

Of course, you get rational coefficients in this case.

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    Thanks! That settles my curiosity :)2012-06-18