Please let me know if I have done everything correctly, especially 2nd paragraph and then I need help and explanation on the last paragraph. TIA
Data Months (x) Population P(x) 0 5 1 10 2 20 3 40 This exponential function measures growth monthly. Using complete sentences, how would I explain how to find the rate of growth every week. Originally, I thought I would calculate the Months (x) out to 12, which would be 20,480 and then divide that by 52, but the more I think about it, the more I am unconvinced that I am right. Next, I have to calculate when the population will reach 500, which would be [\frac{ 500 }{ 5 }=\frac{ 5(2^{x}) }{ 5 }\]\[100=2^{x}\]log(100) = (log)2^2\ log(100) = x (log)2 log(100)/log(2) = x x = 6.64385619 Last, the system I have to put the function into only works in logarithms of base 10. Using complete sentences, how would I explain converting my exponential function P(x) in a logarithmic one and then into a base 10 logarithm. This one I need help on.
\[\frac{ 500 }{ 5 }=\frac{ 5(2^{x}) }{ 5 }\]
Sorry... that what the equation should have been
Looking at the data we notice the population doubles every month. So the exponential function is 2^x. P(x) = A * 2^x when x = 0, P = 5 5 = A * 2^0 = A * 1. So A = 5 P(x) = 5 * 2^x where P is the population in month x.
If we want the Population as a function of w (where w represents week) then: P(w) = 5 * 2^(w/4) assuming 4 weeks in a month.
P(x) = 5 * 2^x When will the population reach 500? 500 = 5 * 2^x 2^x = 100 Take logarithm to the base 10 on both sides: x * log(2) = log(100) = 2 x = 2 / log(2) = 6.64 months
I started there, but because there really isn't 4 weeks in a month, I went to the 52 weeks
So I was correct on that part
The 6.64 months
Yes. The population will reach 500 in 6.64 months.
Okay, but what about the rate of growth every week. Wouldn't you have to calculate the growth at month 12 and then divide by 52 weeks?
This month to week conversion is an approximate one because months have: 28-31 days and the growth in month February will not be the same as the growth in March. But going the year route may be better than the 4 weeks to a month route.
So would I be correct in my answer?
let me see. 52 weeks = 12 months 1 week = 12 / 52 months = 3/13 month P(x) = 5 * 2^x (where x is in months) OR P9w) = 5 * 2^(3w/13) (where w is in weeks)
P(w) = 5 * 2^(3w/13) (where w is in weeks)
Instead of dividing w by 4 to convert it to a month we are dividing w by 4.33333 to convert it to an average month.
1 week is 3/13th of a month w weeks will be (3w/13)th of a month P(x) = 5 * 2^x OR P(w) = 5 * 2^(3w/13)
For the last part, P = 5 * 2^x To convert an exponential function to a logarithmic function, we take log (to the base 10) on both sides of the equation: log(P) = log(5*2^x) = log(5) + log(2^x) = log(5) + x * log(2) log(P) = log(2) * x + log(5) If you want you can substitute decimal values of log(2) and log(5) or you can just leave it the way it is.
Ok, I understand that but could it be done the way I was thinking?
Thank you for the last part.
For the middle part, they actually want you to convert the function so that instead of population being a function of x (the month) they want the population to be a function of w (the week). So if I want to know what the population is in week 5, I should be able to put w = 5 in the formula and calculate P. The way you are doing, you are calculating the population after 1 year then dividing that by 52. That will be the averaging a years population growth equally among 52 weeks. That would assume a constant rate of growth week after week whereas this is an exponential growth.
If the question had been: Find the AVERAGE rate of growth per week (averaged over a one year period) then your method is correct.
Just noticed they want the RATE of growth, not just the population growth as a function of week! The general exponential growth function is: y = a * (1 + r)^x, where a is the initial value and r is the growth rate. P(w) = 5 * 2^(3w/13) P(w) = 5 * (2^(3/13))^w P(w) = 5 * 1.17346^w P(w) = 5 * (1 + 0.17346)^w 0.17346 or 17.346% is the growth rate.
@ranga ... thank you so much... again... just got back to my computer and saw the rest of your post.
You are welcome.
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