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
eq 0 ) is a constant that represents the gro
( n(r)=n_0e^{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 1700 after 1 day, what is the size of the colony after 4 days?
approximately 3842 mosquitoes
(do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Find the value of \( k \)

Given \( N(t)=N_0e^{kt} \), when \( t = 0 \), \( N(0)=N_0 = 1000 \). When \( t = 1 \), \( N(1)=1700 \). Substitute into the formula: \( 1700=1000e^{k\times1} \), then \( e^{k}=\frac{1700}{1000}=1.7 \), so \( k=\ln(1.7) \).

Step2: Find \( N(4) \)

Now that \( N_0 = 1000 \) and \( k=\ln(1.7) \), for \( t = 4 \), \( N(4)=1000e^{4\ln(1.7)} \). Using the property \( a\ln(b)=\ln(b^a) \), we have \( N(4)=1000e^{\ln(1.7^{4})} \). Since \( e^{\ln(x)}=x \), then \( N(4)=1000\times1.7^{4} \). Calculate \( 1.7^{4}=1.7\times1.7\times1.7\times1.7 = 8.3521 \). So \( N(4)=1000\times8.3521 = 8352.1\approx8352 \).

Answer:

The size of the colony after 4 days is \( 8352 \) mosquitoes.