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The population of Lubbock, TX was 149,000 in 1970 and 174,000 in 1980 .(a) Assuming exponential growth, find the function $P(t)$ that describes the population of Lubbock, letting 1970 be time zero. (b) In what year will the population of Lubbock reach $300,000 ?$

(a) $P(t)=149000 e^{0.0155109 t}$(b) 2016

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 7

Applications of Exponential and Logarithmic Functions

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

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Population Growth The popu…

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If we're given that the population in 1970 was 149,000. And that the population in 1980 was 174,000, what is the exponential function to describe this relationship? So to start, what we need to find is that growth rate. So we're going to be using our exponential growth formula on the right hand side and blue here and we can just plug in the information that we do have. So we're gonna start with this population in 1980, which was 174,000, Setting that equal to our initial population in 1970. 149,000 times e to the power of our rate. That's what we're trying to find and that was a period of 10 years. So 174,000, divided by 149,000 here gives us one 167785. Equal to E. to the power of our rate, times 10. Let's take the natural log of both sides and then divide by 10. Which gives us a rate That is equal to 0.01 55 one oh nine. So that rate is important for us to remember. And now let's go ahead and just plug this into a very generic formula for this population at time. T. This gives us population at time T. Then equal to that initial population of 149,000 Times E. To our rate which is we just found 0.0155109 times T. And that gives us that very generic formula here that we can use with any time. And now for part B We want to find at which time or which year was the population equal to 300,000? So now we're going to use this formula we just found let's take the population when it's 300,000. That's what we want to know. This is P. At time T whatever time T. Is Equal to 149,000 Times E. to our growth rate 0.0155109 Times T. Let's go ahead and divide both sides by 149,000. So we get 2.013 is then equal to E. to the power of 0.0155109. The power of T. Let's take the natural log of both sides. And then we can just let's just divide both sides here by that growth rate 0.0155109. And that will give us tea. So we can see here that our time Is then equal to 46. If the time is equal to 46, this is 46 years right here and we know that our initial time or p note was 1970. Let's take 1970 plus those 46 years. And that gives us a year when the population is equal to 300,000 of 2016. Perfect. And then now, just so you know, I did round a little bit in here, this 2.013 and these growth rates around it a little bit in my calculations. So if your number is slightly off, that's probably why, however, it's still fairly accurate.

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