Question 201674
{{{A = A[0]e^(kt)}}}
First we should understand what {{{A[0]}}} stands for and what number it is in your problem. Think about what happens at the start. In other words think about what happens when t=0. If t=0, then the exponent, kt, is also 0 since anything times 0 is 0. And {{{e^0=1}}}. And so our equation. when t=0, is
{{{A = A[0]*1 = A[0]}}}
This tells us that {{{A[0]}}} is the initial population. In your problem we want to "start" the heard at 500 and then figure out the population after 5 years. So we will use {{{A[0] = 500}}}.<br>
Before we can figure out the answer we will also need to know what "k" is. To find "k" we can use an ordered pair that we know fits this equation. We know that when t = 11  the population will have tripled. And if the population started at 500, tripled would be 1500. So A = 1500 when t - 11. Substituting these (and {{{A[0]}}}) into the equation we get:
{{{1500 = 500e^(k(11))}}}
or
{{{1500 = 500e^(11k)}}}
We will now solve for "k". Dividing by 500 we get:
{{{3 = e^(11k)}}}
Finding the natural log of both sides:
{{{ln(3) = ln(e^(11k))}}}
We can get ln(3) from a calculator. The right side, if you understand logarithms, is simply 11k.
{{{1.0986122886681097 = 11k}}}
Dividing by 11
{{{0.0998738444243736 = k}}}
(You may want to round this number. It will make the rest of the problem easier to work with but also make your final answer a little less accurate. I will continue to use this long decimal.)<br>
Now we have the equation:
{{{A = 500e^(0.0998738444243736*t)}}}
To find the population after 5 years we just substitute 5 for t:
{{{A = 500e^(0.0998738444243736*5)}}}
{{{A = 500e^(0.4993692221218680)}}}
We can find {{{e^(0.4993692221218680)}}} from a calculator:
{{{A = 500*1.6476816217236238}}}
Multiplying
{{{A = 823.8408108618119200}}}
So the deer population, after 5 years, is approximately 824 (no fractional deer!?)