The question is if the population takes 20 years to triple, how many years does it take to double.
$\displaystyle \displaystyle y'=Ce^{k20}$
We know that the population in 20 years ($\displaystyle y_{20}$) is triple the starting one ($\displaystyle y_0$):
$\displaystyle y_{20}=3y_0=y_0 e^{20k}$
It follows that
$\displaystyle e^{20k}=3$
and
$\displaystyle k=\frac{\ln(3)}{20}$.
Now set up equation for doubling in unknown $\displaystyle x$ time:
$\displaystyle y_x=y_0 e^{x\frac{\ln(3)}{20}}=2y_0$
and get from that
$\displaystyle e^{x\frac{\ln(3)}{20}}=2$
$\displaystyle x\frac{\ln(3)}{20}}=\ln(2)$
$\displaystyle x=20\frac{\ln(2)}{\ln(3)}=12.62$
Okay, then a more appropriate equation would be:
$\displaystyle y = y_oe^{kt}$
Then you say, when t = 20 years, the population y is 3 times that of the initial population y_o
$\displaystyle 3y_o = y_oe^{k(20)}$
This gives you $\displaystyle e^{20k} = 3$
And then you continue like courteous did.
Exponentials to all bases and logarithms to all bases are interchangable. I prefer to avoid "e" in favor of an obvious base.
Saying that a population take "T" years to triple is the same as saying it is given by $\displaystyle P(t)= C3^{t/T}$ since "t/T" is the number of times it triples. So asking for the number of years to double is asking form t such that $\displaystyle C3^{t/T}= 2C$. Divide both sides by C and take the logarithm (to any convenient base) of both sides: (t/T)log(3)= log(2), t= T (log(2)/log(3)).
Here, T= 20 so the time to double is 20 (log(2)/log(3)).