SOLUTION: 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 e

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 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 e      Log On


   



Question 341314: 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!

Answer by nerdybill(7384) About Me  (Show Source):
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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?
.
Yes, you should be using
P(t) = P0e^kt
.
From the problem you are given:
k = 4.4% = .044
.
Let x = Po (initial population)
then
2x = population doubles
.
So, instead of:
P(t) = P0e^kt
We now have:
2x = xe^(.044t)
.
Now, we solve for t:
2x = xe^(.044t)
dividing both sides by x:
2 = e^(.044t)
take ln of both sides:
ln(2) = .044t
ln(2)/.044 = t
15.75 years = t (doubling time)