Representing any number in $R_+$ by means of two numbers I have the following question:
Let $R_+$ denote the nonnegative reals. Let $0<a<1$ and $b>0$, and set $p = a^n b^m$. Is it possible to find a $n,m\in N$ such that $p$ can approximate any number in $R_+$ arbitrarily close?

I give here a proof that it's possible if either $\frac{\ln(a)}{\ln(b)}$ or $\frac{\ln(b)}{\ln(a)}$ is a normal number (and $b>1$) :
https://en.wikipedia.org/wiki/Normal_number
First notice that there is equivalence between "$p(n,m)$ can approximate any number in $R_+$ arbitrarily close", and "$\ln(p(n,m))$ can approximate any number in $R$" (this is a consequence of the continuity of $\ln$)
$$\ln(p(n,m)) = n\ln(a) + m\ln(b)$$
Now take an arbitrary $x \in R$. If $\gamma = \frac{\ln(a)}{\ln(b)} < 0$ is a normal number, then the k-th first decimal of $\frac{x}{\ln(b)}$ are the decimal of $\gamma$ between $n_k$ and $n_k+k$ for a certain $n_k$ that exist by normality of $\gamma$. You have then :
$$  -10^{-k} < 10^{n_k} \gamma - \lfloor 10^{n_k} \gamma \rfloor +\lfloor \frac{x}{\ln(b)} \rfloor - \frac{x}{\ln(b)}  < 10^{-k}$$
Multiply this by $\ln(b)$, and you get 
$$  -\ln(b) 10^{-k} <  10^{n_k} \ln(a) + \left( - \lfloor 10^{n_k} \gamma \rfloor + \lfloor \frac{x}{\ln(b)} \rfloor \right) \ln(b) - x   < \ln(b) 10^{-k}$$
Now, notice that $m=- \lfloor 10^{n_k} \gamma \rfloor + \lfloor \frac{x}{\ln(b)} \rfloor $ is a positive integer if $n_k$ is taken big enough, and this is possible.
So you indeed have
$$\left| n \ln(a) + m\ln(b) - x \right| < \ln(b) 10^{-k}$$
And x can be approximed arbitrarily close.
Now, notice that I only proved that it's a sufficent condition, not a necessarily one.
