1. ## Population of Elk

I am having trouble getting this one started...

The population of a herd of elkwas 220 in 1970 and 530 in 1990. What is the population today?

2. Hello, jigger1!

The population of a herd of elk was 220 in 1970 and 530 in 1990.
What is the population today?
With a population problem, we assume an exponential function: . $P \;=\;Ae^{Bt}$

In 1970 $(t=0),\;P = 220$
. . We have:. $220 \:=\:Ae^0\quad\Rightarrow\quad A \,=\,220$

The function (so far) is: . $P \;=\;220e^{Bt}$

In 1990 $(t = 20),\;P = 530$
. . We have: . $530 \:=\:220e^{20B} \quad\Rightarrow\quad e^{20B} \:=\:\frac{53}{22}$

Take logs: . $\ln\left(e^{20B}\right) \:=\:\ln\left(\frac{53}{22}\right) \quad\Rightarrow\quad 20B\!\cdot\ln e \:=\:\ln\left(\frac{53}{22}\right)$

. . $20B \:=\:\ln\left(\frac{53}{22}\right)\quad\Rightarrow \quad B \:=\:\frac{1}{20}\ln\left(\frac{53}{22}\right) \;\approx\; 0.044$

Hence, the function is: . $P \;=\;220e^{0.044t}$

Today $(t = 38)$, we have: . $P \;=\;220e^{(0.044)(38)} \;=\;1171.016608 \;\approx\;\boxed{1171\text{ elk}}$

3. Thanks Soroban, how did you know that it would be an exponential function equation since it was a population problem? Can you explain?

I think that is half my problem...