# Thread: exponential growth

1. ## exponential growth

The question is if the population takes 20 years to triple, how many years does it take to double.

$\displaystyle y'=Ce^{k20}$

2. We know that the population in 20 years ( $y_{20}$) is triple the starting one ( $y_0$):
$y_{20}=3y_0=y_0 e^{20k}$

It follows that
$e^{20k}=3$
and
$k=\frac{\ln(3)}{20}$.

Now set up equation for doubling in unknown $x$ time:
$y_x=y_0 e^{x\frac{\ln(3)}{20}}=2y_0$

and get from that

$e^{x\frac{\ln(3)}{20}}=2$

$x\frac{\ln(3)}{20}}=\ln(2)$

$x=20\frac{\ln(2)}{\ln(3)}=12.62$

3. Is this the actual equation given to you?

Does y' represent a derivative here?

EDIT: Is k a constant or is it the time in years?

4. That is the actual question, and k is a constant. The equation was made by me from my interpretation.

5. Okay, then a more appropriate equation would be:

$y = y_oe^{kt}$

Then you say, when t = 20 years, the population y is 3 times that of the initial population y_o

$3y_o = y_oe^{k(20)}$

This gives you $e^{20k} = 3$

And then you continue like courteous did.

6. 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 $P(t)= C3^{t/T}$ 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 $C3^{t/T}= 2C$. 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)).