Problem involving recurrence equation I have a problem involving two recurrence equations and I can't find an algebraic solution for it. I can however use Excel to determine its solution by generating their terms and check when their difference goes to zero.
The problem

The population growth in city A is described by:

$a_1=120000$
$a_{n+1}=a_n+0.02(1-\frac{a_n}{230000})\times{a_n}$

and in the city B is described by:

$b_1=250000$
$b_{n+1}=b_n+0.03(1-\frac{b_n}{140000})\times{b_n}$

Is there a 'n' such that both cities have the same number of population?

If I could express both series in terms of n instead of in terms of previous term then the solution would be to determine that n such as $a_{n+1}=b_{n+1}$ but both series are polynomial and the whole expression becomes ugly and complicated.
With the help of Excel I could determine that for n=39 they have about the same number of population (not exactly but just about).
Question : Can this problem be solved only with a pen and paper, i.e. without computation tools?
Note: this is a high-school problem but I would accept any advanced solution (i.e. college/university level).
PS: I've assumed that if there exists a n such that $a_{n+1}=b_{n+1}$ then perhaps for that n the $a_n=b_b$ and I've tried to determine that $a_n$ by it led nowhere.
 A: Note that $23$ and $7$ are primes. We show by induction that $\displaystyle a_n=\frac{u_n}{(23)^{v_n} w_n}$, with $u_n, w_n, v_n \in \mathbb{Z}$, $v_n\geq 0$, $23$ not dividing $
u_nw_n$, and $7$ not dividing $w_n$ (in fact the only possible prime factors of $w_n$ are $2$ and $5$). This is true for $n=1$ with $v_1=0, w_1=1$, $u_1=120000$. If true for $n$, we have $\displaystyle a_{n+1}=\frac{u_{n+1}}{(23)^{v_{n+1}} w_{n+1}}$ with
$$u_{n+1}=102.10^4u_nw_n(23)^{1+v_n}-2u_n^2 $$
$$w_{n+1}=10^6 w_n^2 $$
and $v_{n+1}=2v_n+1$. One immediately see that $23$ do not divide $u_{n+1}w_{n+1}$ and $7$ do not divide $w_{n+1}$. As $v_1=0$, one has $v_n=2^{n-1}-1$. 
We also show that $\displaystyle b_n=\frac{r_n}{(7)^{s_n} t_n}$ $r_n, s_n, t_n \in \mathbb{Z}$, $s_n\geq 0$, with $7$ not dividing $
r_nt_n$, and $23$ not dividing $t_n$, in the same manner. 
Now we show that $a_n=b_m$ has no solutions.  If $a_n=b_m$, we have:
$$ u_n 7^{2^{m-1}-1}t_m=r_mw_n(23)^{2^{n-1}-1} $$
If $n\geq 2$, then $23$ divide the second term, and not the first. If $m\geq 2$, $7$ divide the first term, and not the second. Hence we must have $n=m=1$, but $a_1=b_1$ is not true. 
