Question

Every 10 years, the Bureau of the Census counts the # of people living in the U.S. In 1790, the population of the U.S. was 3.93 million. By 1800, this # had grown to 5.31 million.

1. Write an exponential function that could be used to model the U.S. population y in millions for 1790 to 1800. Write the equation in terms of x, the number of decades since 1790.

Answer 1

What do exponential functions look like again?
P=Ae^kt?

Answer 2

Would it be something like this?\[P_n=P_0(1.33)^n\]
where P is the population and n is the number of decades? This is probably way off, it's been a while.

Answer 3

Oh, nevermind. I think were both wrong. Is it by any f(x)= ?

Answer 4

That would make 1820 somewhere around 9 million? No?

Answer 5

Exponential function: y = a times b^x

Answer 6

Yea, a is the initial population, the one in 1890. b is 1.33, and x is the number of decades, n in the one I suggested.

Answer 7

But is it way off?

Answer 8

@mathmate please help
@escolas I am not even for sure, I hate world application problems.

Answer 9

The question makes it sound like you should have answers somewhere for 1820 and 1840. Do you have that information?

Answer 10

I don't even get the problem

Answer 11

Is there a table somewhere? Or any further information?

Answer 12

Do you have an example somewhere of a problem like this?That would help in telling me which formula i should apply to this!

Answer 13

I think my number is right. Do you understand exponential functions?

Answer 14

I think i got the answer to the first question but haven't had the same luck for the second.
5.31 = 3.93e^x
Taking ln both side
ln(5.31) = ln(3.93) + x ln(e)
ln(5.31) - ln(3.93) = x where ln(e) =1
x = 0.3009524093726 ...........................Ans

Answer 15

Yes, I understand Exponential functions, but just not in a word problem; and I already did #2 of the problem already.

Answer 16

@escolas Does the way i did it make sense?

Answer 17

I thought it was f(x) = A * B^X

Answer 18

No @Najia2000, I think you want to use exponential unctions, not logarithmic ones.

Answer 19

My equation should be right. Give me a minute, I'll try to work out a written solution to be clearer.

Answer 20

Ohh, okay.
@escolas
Sorry if I caused any confusion!

Answer 21

This might go off topic of the problem, but what's y > 0 in interval notation

Answer 22

In range ...?

Answer 23

yes

Answer 24

So we want to let x be the number of decades since the initial data in 1790. Since we know what the population is in both 1790 and 1800, we can compare and use an x value of 1 to represent the 1 decade that has passed. Plugging in what we know, we have\[f(x)=a*b^x \rightarrow 5.31=3.93*(something)^1\]Now we can solve for that something by dividing by both sides. We get that something is 5.31/3.93 which is somewhere around 1.33. Then our equation is \[f(x)=P*1.33^x\] where P is the initial population, 3.93 and x is the number of decades since 1790.
Does this make sense?

Answer 25

y > 0 is (0, infinity)

Answer 26

Since it's strictly greater than it gets parenthesis on the left, and infinity always gets a parenthesis as well.

Answer 27

@escolas I think you deserve infinity medals. :p

Answer 28

thank you so much @escolas

Answer 29

Haha I just hope it's clear. Not a problem at all!

Answer 30

@escolas If he had to use logarithm functions would the way i did it be correct?

Answer 31

Yea hehe, there's two parts to #1

Answer 32

I'm not sure @Najia2000. I don't know if you can really model population logarithmically.

Answer 33

@escolas I need help with #1 the equation portion

Answer 34

Didn't we do that? It was \[y=3.93*1.33^x\]

Answer 35

that exponential function isn't it

Answer 36

If you replace 1.33 with 1.35115 it is.

Answer 37

@satellite73 would the answer just be y = 3.93(1.35)^x or should there be 2 answers?