(a) If $\mathrm{N}$ is the population of the colony and $\mathrm{t}$ is the time in days, express $\mathrm{N}$ as a function of $\mathrm{t}$. Consider $\mathrm{N}_{0}$ is the original amount at $t=0$ and $k \neq 0$ is a constant that represents the growth rate.
\[
\mathrm{N}(\mathrm{t})=\mathrm{N}_{0} e^{\mathrm{kt}}
\]
(Type an expression using $t$ as the variable and in terms of $e$.)
(b) The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there are 1800 after 1 day, what is the size of the colony after 4 days?

Approximately $\square$ mosquitoes
(Do not round until the final answer. Then round to the nearest whole number as needed.)

Step 1 ：The population \(N\) of the colony as a function of time \(t\) is given by the exponential growth formula: \[N(t) = N0 * e^{kt}\] where: \(N0\) is the initial population size, \(k\) is the growth rate, \(t\) is the time in days, and \(e\) is the base of the natural logarithm (approximately equal to 2.71828).

Step 2 ：To find the size of the colony after 4 days, we first need to find the growth rate \(k\). We know that the initial population \(N0\) is 1000, and the population after 1 day (\(N(1)\)) is 1800. We can substitute these values into the exponential growth formula and solve for \(k\): \[1800 = 1000 * e^k\] => \[e^k = 1800/1000 = 1.8\] => \[k = \ln(1.8)\] (taking natural logarithm on both sides).

Step 3 ：Now that we have the growth rate \(k\), we can find the population after 4 days (\(N(4)\)) by substituting \(N0 = 1000\), \(k = \ln(1.8)\), and \(t = 4\) into the exponential growth formula: \[N(4) = 1000 * e^{4*\ln(1.8)}\] => \[N(4) = 1000 * (1.8)^4\] => \[N(4) = 1000 * 10.4976\] => \[N(4) = 10498\]

Step 4 ：So, the size of the colony after 4 days is approximately 10498 mosquitoes, rounded to the nearest whole number. Therefore, the final answer is \(\boxed{10498}\).