"A model should be as simple as possible but no simpler..."
Albert Einstein
$$\begin{eqnarray} N(t = 0) &=& N_0 \ &=& 7\ 000\ 000\ 000 \end{eqnarray}$$
$$N(t + 1) = \lambda \times N(t) + N(t)$$ - where $\lambda = 0.015$
$$N(t + 1) = (\lambda / 12) \times N(t) + N(t)$$
$$N(t + 1) = \lambda \times N(t) + N(t)$$
$$N(t + 1) = \lambda \times N(t) + N(t)$$
$$ \mathit{i.e.\quad} N(t + 1) = (\lambda + 1) \times N(t)$$
$$\begin{eqnarray} N(t + 2) &=& (\lambda + 1) \times N(t + 1) \ &=& (\lambda + 1)\times(\lambda + 1) \times N(t) \ &=& (\lambda + 1)^2 \times N(t) \end{eqnarray}$$
$$N(t + n) = (\lambda + 1)^n \times N(t)$$
So, the number of years ($n$) taken to double is the $n$ that satisfies
$$(\lambda + 1)^n = 2$$
$$\begin{eqnarray} (\lambda + 1)^n &=& 2 \ \log((\lambda + 1)^n) &=& \log(2) \ n \times \log(\lambda + 1) &=& \log(2) \ n &=& \log(2) / \log(\lambda + 1) \ n &=& 46.6 \end{eqnarray}$$
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