HELP!! Can Some1 Please help me with these polulation problems?

Hello again, rom!

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2. Population growth

The population of a certain city was 30,000 in 1990 and 45,000 in 2000.
The growth of this population can be described by the formula: $P \;=\;P_oe^{kt}$
. . where $P_o$ is the population size at a particular time $t$ , and $k$ is a constant.

(a) Use this information to find the value of $k.$

In 1990 $(t = 0),\;P = 30,000$
. . So we have: . $30,000 \:=\:P_oe^0\quad\Rightarrow\quad P_o = 30,000$
The function (so far) is: . $P \;=\;30,000e^{kt}$

In 2000 $(t=10),\;P = 45.000$
. . So we have: . $45,000 \:=\:30,000e^{10k}\quad\Rightarrow\quad e^{10k} \:=\:1.5$

Take logs: . $\ln\left(e^{10k}\right) \:=\:\ln(1.5)\quad\Rightarrow\quad 10k(\ln e) \:=\:\ln(1.5)$

. . Hence: . $k \:=\:\frac{\ln(1.5)}{10}\:\approx\:0.04$

Therefore, the function is: . $P \;= \;30,000e^{0.04t}$

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(b) Then make a prediction of the population size for the year 2020.

In 2020 $(t = 30)\!:\;\;P \:=\:30,000e^{(0.04)(30)} \:\approx\:99,604$

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(c) Apply a suitable transformation to the population growth formula
so that you obtain an expression of the form: $y \:=\:mx=+c$

We have: . $P \:=\:30,000e^{0.04t}$

Take logs:
. . $\ln(P) \:=\:\ln\left(30,000e^{0.04t}\right) \:= \:\ln(30,000) + \ln\left(e^{0.04t}\right) \:=\:\ln(30,000) + 0.4t(\ln e)$

And we have: . . . . $\underbrace{\ln(P)}_\downarrow \;= \;\underbrace{0.4t}_\downarrow + \underbrace{\ln(30,000)}_\downarrow$
which is of the form: . $y \;\;\;= \;\;\;mx \quad + \quad c$

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(d) How do $x,\,y,\,m$ and $c$ relate to the variables $P$ and $t$ ,
and the constants $P_o$ and $k$ in the original formula?

. . $\begin{array}{cccc}x \\ y \\ m \\ c\end{array}<br /> \begin{array}{cccc} = \\ = \\ = \\ = \end{array}<br /> \begin{array}{cccc}t \\ \ln(P) \\ k \\ \ln(P_o)\end{array}$