Question

the population of a colony of mosquitoes obeys the law of uninhibited growth. use this information to answer parts (a) through (c).
(a) if n is the population of the colony and t is the time in days, express n as a function of t. consider ( n_0 ) is the original amount at ( t = 0 ) and ( k
eq0 ) is a constant that represents the growth rate.
( n(t)=n_0e^{kt} )
(type an expression using t as the variable and in terms of r.)
(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 3 days?
approximately ( square ) mosquitoes
(do not round until the final answer. then round to the nearest whole number as needed)

Explanation:

Step1: Find the value of \(k\)

We know that \(N(t)=N_{0}e^{kt}\). Given \(N_{0} = 1000\), \(t = 1\) and \(N(1)=1800\).
Substitute these values into the formula: \(1800=1000e^{k\times1}\).
Then \(\frac{1800}{1000}=e^{k}\), so \(e^{k}=1.8\).
Take the natural logarithm of both sides: \(k=\ln(1.8)\).

Step2: Find the population after \(t = 3\) days

Now that we know \(k=\ln(1.8)\) and \(N_{0}=1000\), and we want to find \(N(3)\).
Substitute into the formula \(N(t)=N_{0}e^{kt}\), so \(N(3)=1000e^{3\ln(1.8)}\).
Using the property \(a\ln(b)=\ln(b^{a})\), we have \(N(3)=1000e^{\ln(1.8^{3})}\).
Since \(e^{\ln(x)}=x\), then \(N(3)=1000\times1.8^{3}\).
Calculate \(1.8^{3}=1.8\times1.8\times1.8 = 5.832\).
So \(N(3)=1000\times5.832 = 5832\).

Answer:

\(5832\)