population of a city based on another [hw] I've been struggling with understanding how to solve a particular branch of population problems.  The question reads as follows:

City P had a population of 20,000 in the year 2005.  The population of city P always increases by 11% every 24 years.  City Q's population doubles in two-thirds of the time that city P's population doubles.  In the year 2010, city Q has 90% as many people as city P.  In the year 2027, what will be the population of city Q?

I attempted to begin this problem with the following equations:

$P_P = 20000e^{.11/24}$  

While I'm not sure what is the correct way to begin this problem, I'm really stumped by how I should approach the other parts of this problem.  Please don't use advanced maths; I am just in pre-calculus.
 A: We do what we believe to be the harder part, and leave the rest to you.
If we want to model $P$'s population by a formula of the type you wrote, it should be 
$$P(t)=20000e^{kt}.$$
The population increases by $11$ percent in $24$ years. Thus
$$(20000)(1.11)=20000e^{24k}.$$
Cancel the $20000$, then take the natural logarithm of both sides. We get $$k=\frac{\ln(1.11)}{24}.\tag{1}$$
Now let's deal with the comparison of doubling times. Suppose that the population of city $Q$ satisfies $Q(t)=Ce^{\ell t}$ for some constant $\ell$. 
The doubling time $d_P$ for city $P$ satisfies $2=e^{kd_P}$, so $d_P=\frac{\ln 2}{k}$. Similarly, the doubling time for city $Q$ is $\frac{\ln 2}{\ell}$. 
The doubling time $d_Q$ for $Q$ is two-thirds of the doubling time for $P$. It follows that 
$$\ell =\frac{3}{2}k={\frac{\ln(1.11)}{16}}.\tag{2}.$$
Now we know the crucial constants for $P$ and $Q$. To finish, do this:


*

*Find the population of $P$ in $2010$.

*That will tell you the population of $Q$ in $2010$.   

*To find the population of $Q$ in $2027$, multiply the population in of $Q$ in $2010$ by $e^{{17\ell}}$. 
There are various shortcuts you can use if you have good control of the algebra of exponentials/logarithms. 
