Approximation of $\log(x)$ as a linear combination of $\log(2)$ and $\log(3)$

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 $$2^{-33} 3^{23}=10.959708460955880582332611083984375 \approx 10.96.$$

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.