MHB How does the population of a southern city follow the exponential law?

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The population of a southern city follows the exponential growth model, expressed as N(t) = N_0e^{kt}. Given that the population doubled in 18 months and is currently 10,000, the growth constant k is approximately 0.462098. Using this value, the projected population in two years is approximately 25,198. A suggestion was made to clarify the equation setup for better understanding, but it was acknowledged as a minor point. The calculations and logic presented were deemed correct overall.
karush
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$\tiny{\textbf{6.8.7}}$ Kiaser HS

Population Growth The population of a southern city follows the exponential law
(a) If N is the population of the city and t is the time in years, express N as a function of t.$N(t)=N_0e^{kt}$
(b) If the population doubled in size over an 18-month period and the current population is 10,000, what will
the population be 2 years from now?
$\begin{array}{rl}
2&=e^{k(1.5)} \\
\ln 2&=k \cdot 1.5\\
\dfrac{\ln 2}{1.5}&=k\\
\therefore k&\approx0.462098\\
f(t)&\approx10000e^{0.462098\cdot 2}\\
&\approx 25198
\end{array}$

well anyway hopefully ok :unsure:
typos maybe
 
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I didn't check the numbers but the logic is correct.

-Dan
 
Since you stated your basic equation as
$N(t)= N_oe^{kt}$
I would have preferred that you write
$2N_0= N_0e^{1.5k}$
before you divide both sides by $N_0$ to get
$2= e^{1.5k}$.

But I realize that is being petty!