Asked by Samantha
Hello I need a little help with my homework. I started trying to work it out but I am not sure if it is the correct solution. Please help!
Question:
Population growth. The exponential growth rate of the population of United Arab Emirates is 4.4% per year (one of the highest in the world). What is the doubling time?
Sources: Based on data from U.S. Census Bureau; International Data Base 2007
My Answer: I changed 4.4% to .044^2. My result was .001936 or round to 0.002%.
Is this correct? Should I be using the P(t) = P0e^kt? If so how should I go by doing so? Thanks for your help in advance!
Question:
Population growth. The exponential growth rate of the population of United Arab Emirates is 4.4% per year (one of the highest in the world). What is the doubling time?
Sources: Based on data from U.S. Census Bureau; International Data Base 2007
My Answer: I changed 4.4% to .044^2. My result was .001936 or round to 0.002%.
Is this correct? Should I be using the P(t) = P0e^kt? If so how should I go by doing so? Thanks for your help in advance!
Answers
Answered by
Dr Russ
If the population after year 1 is
P(1)=P(0)x1.044
at the end of year 2 it is
P(2)=P(1)x1.044 = P(0)x1.044^2
at the end of year t it is
P(t)=P(0)x(1.044)^t
to find when population has doubled (i.e. is 2P(0))
2P(0)= P(0)x(1.044)^t
and solve for t
2 = (1.044)^t
log(2)=log(1.044)^t)= txlog(1.044)
thus t=log(2)/log(1.044)
and I got 16.1 years
P(1)=P(0)x1.044
at the end of year 2 it is
P(2)=P(1)x1.044 = P(0)x1.044^2
at the end of year t it is
P(t)=P(0)x(1.044)^t
to find when population has doubled (i.e. is 2P(0))
2P(0)= P(0)x(1.044)^t
and solve for t
2 = (1.044)^t
log(2)=log(1.044)^t)= txlog(1.044)
thus t=log(2)/log(1.044)
and I got 16.1 years
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