Question

Please i really need the help asap! please!!

Do some research and find a city that has experienced population growth.
Determine its population on January 1st of a certain year. Write an
exponential function to represent the city’s population, y, based on the
number of years that pass, x after a period of exponential growth. Describe
the variables and numbers that you used in your equation.
and these are the numbers i have for it
2011 2012 2013 2014
Salem, Or 156,148 157,788 159,138 160,637
detroit, Mi 711,195 702,149 696,746 696,746

Answer 1

i don't expect the answer to be given straight to me i just need some help to be walked through it.

Answer 2

\(\Huge\color{#EB00FF}{\text{WELCOME}}\) \(\Huge\color{blue}{\text{TO}}\) \(\Huge\color{green}{\text{OPEN}}\)\(\Huge\color{purple }{\text{ STUDY!!!!!!!!!!!}}\) \(\Huge\heartsuit\)

Answer 3

thanks! ouo

Answer 4

@mathmate

Answer 5

i cant see some replys for some reason so can you pm when you reply??

Answer 6

Ok im here :PP

Answer 7

ahh thanks @Christopher1448

Answer 8

@koikitty
Have you got help as needed?
If so, please close the question so helpers don't spend time unnecessarily.
Thank you!

Answer 9

@mathmate I havent gotten the help i needed yet, but as soon as i do i will close the question.

Answer 10

Well, the question asks for a city that has population growth. One of the cities you posted has a decline (decay, in math terms). But for me, it is just as good as any for an exercise.
If you still need help to find the exponential equation, here are some hints:
1. When you try to fit an exponential equation to real data, do not expect exact fit, unless you have only three points.
2. The equation you make will work \(approximately\) well between your data points. Do not try to extrapolate, i.e. use the equation for points outside of the range of the independent variable you used.
3. Properties of exponential functions:
If the expected equation is
\(\large y=ab^x\)
and (x1,y1),(x2,y2), (x3,y3) ... are selected data points, set x=0 for the first year in the data points, do not use x=1970, or 1980, etc. Use x=0 for the year 1970, and state that year 0 is 1970. Otherwise you will overload your calculator and get inaccurate results.
From this we can find a, and b.
y2/y1=bx^(x2-x1)
Use logarithm to solve for b.
a=y1/(b^x1) or a=y2/b^x2.... but they won't be equal for the reason I mentioned above

Answer 11

so the exquation would be 156,148^2/711,195=b0^(0^20^1) @mathmate

Answer 12

Can you explain how you got this equation?
Is it in the form of:
\(\large y=ab^x\)
What did you get for a, and for b?

Answer 13

uhm i didnt use y=ab^x i used a=y1/(b^x1) im not really sure how to set it up

Answer 14

Do not confuse the basic equation with particular points on the curve.
y=ab^x is the basic equation.
y1=ab^(x1) and so on are particular points on the curve that you have found from your data.
The object is to use particular points to solve for the constants a and b.
Whichever way you use, you should get something reasonable.
For Salem, I suggest you use 2011 as the base year, then x=0, y=156148 and
for 2014, x=3, y=160637.
Consequently
using
y=ab^x
156148=ab^0
160637=ab^3
Solve for a and b.

This way, you are covering the whole period 2011 to 2014, although discrepancies are bound to arise for the intermediate years.
If you want the best fit, you need to use regression, but I don't know if that's the object of the exercise since the word is not anywhere in the question.

Answer 15

Im not really sure who to solve for a and b..
do i need to get a and b by them selves and the solve for them? @mathmate
im sorry iv always really struggled with math ;~; but i really want to get better

Answer 16

`156148=ab^0` ....(1)
`160637=ab^3` ....(2)
To solve for b, divide equation (2) by equation (1), then
160637/156418=ab^3/(ab^0)=b^3
take cube roots on both sides,
b=(160637/156418)^(1/3) [use your calculator]
=1.00891
Then put b=1.00891 into equation (1) to solve for a.
156418=a(1.00891)^0=a
Since
a=156418,
b=1.00891,
the equation is then
y=156418(1.00891^x)
where x is the number of years since 2011, and
y is the population of year 2011+x.

Check estimates for years 2012 and 2013,
2012: x=2012-2011=1
population (2012)=156418(1.00891^1)=157812 (compared with data = 157788)
I'll let you compare the year 2013.

Answer 17

okay this makes more sence now!
so it would be
2013=159,138(1.00891)^1 and that equals 160555

Answer 18

@koikitty
Sorry, nope.
We have to make a clear distinction between the data set (that you collected) and the equation to estimate the data.
Our equation was established as:
y=156418(1.00891^x) .......................(3)
The only variable that changes is x, number of years since 2011.
y is the \(estimated\) population, which we can compare with the data set. The two should be reasonably close.
To estimate any and all years,
we use a=156418 and b = 1.00891 as shown in the above equation (3).
All we need to do to estimate a year is to calculate x, and substitute into equation (3).
For 2012, we have x=(2012-2011)=1.
So
equation (3) gives
y=156418(1.00891^1) =156418(1.00891)=157812 (estimated value)
actual value is 157788. So the error due to estimation is (157812-157788)/157788=+0.015%, which is quite accurate.

So proceed the same way for 2013, where
x=2013-2011=2
and plug into equation 3 above to get the \(estimate\) of the 2013 population.
You are free to compare with your actual data to see if the estimate is still good.

Answer 19

i got 161,986
y=159,138(1.00891^1)
y=159,138(1.0179)
161,986
aah i hope i got it right.. i understand better on how to do it @mathmate

Answer 20

You may not have noticed. The value of a equals 156148, and the value of b remains at 1.00891 at \(all\) times. If you change them, you are solving a different problem. The only number that changes with time is x.
That is why the function is written
\(y=ab^2\) where a and b are constants above (i.e. numbers that do not change, once found).
So for 2012, it is
P(1)=\(156148(1.00891^{1})\) where 1=2012-2011