Exponential growth is modeled by the equation$\displaystyle A=Pe^{kt}$

where $\displaystyle A$ is the final amount, $\displaystyle P$ is the inital amount, $\displaystyle k$ is the growth constant and $\displaystyle t$ is time.

In this problem, $\displaystyle A=800K$, $\displaystyle P=200K$ and $\displaystyle t=3$.  Substituting these variables into the growth equation the solving for $\displaystyle k$ gives us

$\displaystyle 800,000=200,000e^{3k}$

$\displaystyle \frac{800,000}{200,000}=e^{3k}$

$\displaystyle 4=e^{3k}$

$\displaystyle ln(4)=ln(e^{3k})$

$\displaystyle ln(4)=3k$

$\displaystyle k=\frac{ln(4)}{3}$

The half-life of an isotope is the time it takes for half the isotope to disappear.  Isotopes decay exponentially.

Exponential decay is also modeled by the equation $\displaystyle A=Pe^{kt}$

where $\displaystyle A$ is the final amount, $\displaystyle P$ is the inital amount, $\displaystyle k$ is the growth constant and $\displaystyle t$ is time.

Since half the isotope has disappeared, the final amount $\displaystyle A$ is half the inital amount $\displaystyle P$, or  $\displaystyle \frac{A}{P}=0.5$.

In this problem, $\displaystyle t=5.27yrs$.

Substituting these variables into the exponential equation and solving for $\displaystyle k$ gives us

$\displaystyle A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$\displaystyle 0.5=e^{5.27k}$

$\displaystyle ln(0.5)=ln(e^{5.27k})$

$\displaystyle ln(0.5)=5.27k$

$\displaystyle k=\frac{ln(0.5)}{5.27}$

Exponential growth is modeled by the equation $\displaystyle A=Pe^{kt}$

where $\displaystyle A$ is the final amount, $\displaystyle P$ is the inital amount, $\displaystyle k$ is the growth constant and $\displaystyle t$ is time.

After $\displaystyle 64dys$, the number of cats has doubled, or the final amount $\displaystyle A$ is double the inital amount $\displaystyle P$, or  $\displaystyle \frac{A}{P}=2$.

In this problem, $\displaystyle t=64$.

Substituting these variables into the exponential equation and solving for $\displaystyle k$ gives us

$\displaystyle A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$\displaystyle 2=e^{64k}$

$\displaystyle ln(2)=ln(e^{64k})$

$\displaystyle ln(2)=64k$

$\displaystyle k=\frac{ln(2)}{64}$

To find the number cats after $\displaystyle 1$ year, $\displaystyle P=2$, $\displaystyle k=\frac{ln(2)}{64}$, and $\displaystyle t=365$

$\displaystyle A=Pe^{kt}=2e^{\frac{ln(2)}{64}*365}\approx 104$

The exponential growth is modeled by the equation $\displaystyle y=a(1+r)^{t}$

where $\displaystyle y$ is the final amount, $\displaystyle a$ is the inital amount, $\displaystyle r$ is the growth rate and $\displaystyle t$ is time.

In this problem, $\displaystyle a=460K$, $\displaystyle r=0.37$, and $\displaystyle t=2013-2003=10$.  Substituting these values into the equation gives us

$\displaystyle y=460,000(1+0.37)^{10}=10.7M$

The exponential growth is modeled by the equation $\displaystyle y=a(1+r)^{t}$

where $\displaystyle y$ is the final amount, $\displaystyle a$ is the inital amount, $\displaystyle r$ is the growth rate and $\displaystyle t$ is time.

In this problem, $\displaystyle a=3M$ and $\displaystyle t=2014-2008=6$. $\displaystyle r=-0.15$ because the rate is decreasing.  Substituting these values into the equation gives us

$\displaystyle y=3,000,000(1-0.15)^{6}=1.13M$

The exponential growth is modeled by the equation $\displaystyle y=a(1+r)^{t}$

where $\displaystyle y$ is the final amount, $\displaystyle a$ is the inital amount, $\displaystyle r$ is the growth rate and $\displaystyle t$ is time.

In this problem, $\displaystyle a=\$1500$, $\displaystyle y=\$6500$ and $\displaystyle t=8 yrs$.  Substituting these variables into the growth equation and solving for r gives us

$\displaystyle 6500=1500(1+r)^{8}$

$\displaystyle \frac{13}{3}=(1+r)^{8}$

$\displaystyle \sqrt[8]{\frac{13}{3}}=\sqrt[8]{(1+r)^{8}}$

$\displaystyle \sqrt[8]{\frac{13}{3}}=1+r$

$\displaystyle 1+r=1.2012$

$\displaystyle r=0.2012=20.12\%$

To find direct constant of proportionality $\displaystyle k$, you need to find the slope of the line between the two points at $\displaystyle t=0$ and $\displaystyle t=\pi$. To do so we use the formula:

$\displaystyle f(x)=e^t-cos(2t)$

$\displaystyle k=\frac{f(\pi)-f(0)}{\pi -0}=\frac{(e^\pi-cos(2*\pi))-(e^0-cos(0))}{\pi}=\frac{e^\pi-1}{\pi}$

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$\displaystyle p(t)=p_0e^{kt}$

Where $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased from 9876 to 10381 between 2013 and 2015, we can solve for this constant of proportionality:

$\displaystyle 10381=9876e^{k(2015-2013)}$

$\displaystyle \frac{10381}{9876}=e^{2k}$

$\displaystyle 2k=ln(\frac{10381}{9876})$

$\displaystyle k=\frac{ln(\frac{10381}{9876})}{2}=0.0252$

Using this, we can calculate the expected value from 2015 to 2030:

$\displaystyle P=10381e^{(2030-2015)(0.0252)} \approx 15150$