Question 347820
a) "a" = {{{root(6, 8/3)}}} which is approximately 1.1776...
This makes your equation: {{{P = (2000*(root(6, 8/3))^t)/(4 + (root(6, 8/3))^t)}}}<br>
b) First we find the years it takes the population to reach 1800:
{{{1800 = (2000*(root(6, 8/3))^t)/(4 + (root(6, 8/3))^t)}}}
Multiply both sides by {{{(4 + (root(6, 8/3))^t)}}}:
{{{7200 + 1800(root(6, 8/3))^t = 2000(root(6, 8/3))^t}}}
Subtract {{{1800(root(6, 8/3))^t)}}} from each side:
{{{7200 = 200(root(6, 8/3))^t)}}}
Divide both sides by 200:
{{{36 = (root(6, 8/3))^t)}}}
Find the logarithm of each side, using a base that your calculator "knows":
{{{log((36)) = log(((root(6, 8/3))^t)))}}}
Use the property of logarithms, {{{log(x, (p^q)) = q*log(x, (p))}}}, to move the exponent/variable out in front:
{{{log((36)) = t*log((root(6, 8/3)))}}}
Divide both sides by {{{log((root(6, 8/3)))}}}:
{{{log((36))/log((root(6, 8/3))) = t}}}
Use your calculator for a decimal approximation:
{{{21.9213625253483390 = t}}}
This is how long it takes for the population to reach 1800. For the time it takes for the population to increase from 800 to 1800, just subtract the 6 years it took to reach 800: 151.9213625253483390.<br>
c) To answer this part, take the equation:
{{{P = (2000*(root(6, 8/3))^t)/(4 + (root(6, 8/3))^t)}}}
and multiply the numerator and denominator on the right by: {{{((1/((root(6, 8/3))^t)))/(1/((root(6, 8/3))^t))}}}. (You'll see why when we're done.)
{{{P = ((2000*(root(6, 8/3))^t)/(4 + (root(6, 8/3))^t))*((1/((root(6, 8/3))^t))/(1/((root(6, 8/3))^t)))}}}
Multiplying the numerators and using the Distributive Property on the denominators we get:
{{{P = 2000/(4/((root(6, 8/3))^t) + 1)}}}
With the equation in this form and with some knowledge about how exponents and fractions work we can figure out why the deer population will never exceed 2000:<ul><li>Since t is time it can never be negative. The lowest value it can be is zero. When t = 0, {{{P = 2000/(4/((root(6, 8/3))^0) + 1)}}}. Since any nonzero number to the zero power is 1 we get: {{{P = 2000/(4/1 + 1) = 2000/5 = 400}}}. So the population at t = 0 (IOW the initial population) is 400.</li><li>As t increases, {{{(root(6, 8/3))^t}}} increases</li><li>As  {{{(root(6, 8/3))^t}}} increases,  {{{4/(root(6, 8/3))^t}}} decreases (because as the denominator increases the value of the entire fraction decreases)</li><li>As {{{4/(root(6, 8/3))^t}}} decreases, {{{4/((root(6, 8/3))^t) + 1}}} decreases</li><li>As {{{4/((root(6, 8/3))^t) + 1}}} decreases, {{{2000/(4/((root(6, 8/3))^t) + 1)}}} increases (because as a denominator decreases, the value of the entire fraction increases).</li><li>So the highest population will come with the highest value of t.</li><li>For very large values of t, the denominator of {{{4/((root(6, 8/3))^t)}}} will become an extremely large positive number. And a fraction with a positive numerator and an extremely large positive denominator will be a positive fraction which is very, very close to zero.</li><li>If {{{4/(root(6, 8/3))^t}}} is a positive number very close to zero then {{{4/((root(6, 8/3))^t) + 1}}} will be a number just a tiny bit above 1.</li><li>And if {{{4/((root(6, 8/3))^t) + 1}}} is a number just barely above 1, then {{{2000/(4/((root(6, 8/3))^t) + 1)}}} is a number just barely <i>below</i> 2000.</li></ul>
This is why the population will never exceed 2000. (In fact, since {{{4/(root(6, 8/3))^t}}} can never actually be zero, {{{4/((root(6, 8/3))^t) + 1}}} can never actually be 1 and the population can never actually be 2000.)