# The bat population in a certain Midwestern county was 230,000 in 2009, and the observed doubling time for the population is 22 years. How do you find an exponential model?

Jan 12, 2017

$B = 230 , 000 \cdot {2}^{\frac{t}{22}}$

$B = 230 , 000 {e}^{\frac{t}{22} \ln 2}$

#### Explanation:

let's say a quantity $Q$, starting with initial value ${Q}_{o}$, doubles over a period $\tau$, we can say that:

$Q = {Q}_{o} \cdot {2}^{\frac{t}{\tau}}$

so checking that:

• for $t = \tau$, $Q = {Q}_{o} \cdot 2$
• for $t = 2 \tau$, $Q = {Q}_{o} \cdot {2}^{2}$
• and so on

So for the bats we can say that:

$B = 230 , 000 \cdot {2}^{\frac{t}{22}}$

If you want that in a calculus friendly form, you can instead start with the general idea of exponential growth/decay:

$\frac{\mathrm{dQ}}{\mathrm{dt}} = \lambda Q$

${\int}_{{Q}_{o}}^{Q} \frac{d Q}{Q} = {\int}_{0}^{t} \lambda \mathrm{dt}$

$Q = {Q}_{o} {e}^{\lambda t}$

If we know that the amount doubles over a period $\setminus \tau$ then:

$\frac{Q}{Q} _ o = 2 = {e}^{\lambda \tau}$

so $\lambda = \frac{1}{\tau} \ln 2$

And our relationship is:
$Q = {Q}_{o} {e}^{\frac{t}{\tau} \ln 2}$

In that case, for the bats we say that:

$B = 230 , 000 {e}^{\frac{t}{22} \ln 2}$