Question

The population of California was 10,586,223 in 1950 and 23,668,562 in 1980. Assume that the population grows exponentially.
(a) Find a function that models the population t years after 1950. (Round rate r to four decimal places.)
(b) Find the time required for the population to double. (Round your answer to the nearest whole number.)
(c) Use the function from part (a) to predict the population of California in the year 2003.

Answer 1

nice numbers...

Answer 2

yeah lol

Answer 3

easy way or hard way?

Answer 4

Whatever you want to do. I just want the answers

Answer 5

probably need the hard way , probably your teacher wants
\[P(t)=P_0e^{rt}\]

Answer 6

yeah that is what he wants

Answer 7

ok first we do it the hard way. then if you want i can show you and easier way, or we can just be done. set
\[23,668,562=10,586,223 e^{30r}\] and solve for r. is this step clear?

Answer 8

that is an o and not a zero, correct? Yes that is clear

Answer 9

it is a zero. 30 years. we just don't know r yet, so we have to find it

Answer 10

first step is to divide, then take the log, then divide by 30 to get r

Answer 11

ok just wanted to make sure. ok

Answer 12

\[2.2358=e^{30t}\]
\[\ln(2.2358)=30t\]
\[t=\frac{\ln(2.2358)}{30}\]

Answer 13

t=.0268 rounded

steps clear so far?

Answer 14

oh typos sorry. we are solving for r, not t !

Answer 15

r = .0268 rounded. we are finding the rate

Answer 16

ok, and I got .0268199702, so then as my final answer i got 23668677.39

Answer 17

\[2.2358=e^{30r}\]

\[\ln(2.2358)=30r\]
\[r=\frac{\ln(2.2358)}{30}\]
\[r=.0268\]

Answer 18

not sure what you mean final answer

Answer 19

the answer to a) is
\[P(t)=10,586,223 e^{.0268 t}\]

Answer 20

the answer to b we still have to compute.

Answer 21

the answer to c) is what you get when you replace t by 53

Answer 22

i mean the answer that i get when i plug in .0268 to \[10586223e^(30)(.0268)\]

Answer 23

yes but you know that already. you already have you population in 1980

Answer 24

oh ok i see

Answer 25

for part c just compute
\[10,586,223e^{53\times .0268}\] and see what you get

Answer 26

43860399.85

Answer 27

for part b) you want doubling time. put
\[2=e^{.0268t}\] and solve for t

Answer 28

you get
\[\ln(2)=.0268t\]
\[t=\frac{\ln(2)}{.0268}\]

Answer 29

or 25.86 so doubles about every 26 years or so

Answer 30

steps clear?

Answer 31

yeah thanks

Answer 32

yw