Question

Growth Population.
http://prntscr.com/d9fab1

Answer 1

2 equations 2 unknowns u can solve for them

Answer 2

ln (p1/A) = kt1
ln (p2/A) = kt2

simplfies to
ln p1- ln A = k*t1
ln p2 - ln A = k*t2

lets say B= lnA
now its in the form
k1 - B = k*t1
k2-B=k*t2

k1,k2,t1,t2 are known solve for B and k

Answer 3

3mar did you get an answer yet?

Answer 4

Can you show me how to, please @wolf1728?

Answer 5

\[222=Ae ^{1992 k}~...(1)\]
\[224=Ae ^{2001 k} ~...(2)\]
\[\frac{ 224 }{ 222 }=e ^{2001k-19992k}=e ^{9k}\]
\[k=\frac{ 1 }{ 9 }\ln \frac{ 224 }{ 222 }\]
from (1)
\[222=Ae ^{1992\left( \frac{ 1 }{ 9 }\ln \frac{ 224 }{ 222 } \right)}\]
\[\ln 222=\ln A+\frac{ 1992 }{ 9 }\left( \ln 224-\ln 222 \right)\]
\[\ln A=\left( 1+\frac{ 1992 }{ 9 } \right)\ln 222-\frac{ 1992 }{ 9 }\ln 224\]
\[\ln A=\frac{ 2001 }{ 9 }\ln 222-\frac{ 1992 }{ 9 }\ln 224=\frac{ 2001\ln 222-1992\ln 224 }{ 9 }\]
\[A=e ^{\frac{ 2001\ln 222-1992\ln 224 }{ 9 }}\]
\[P=e ^{\frac{ 2001\ln 222-1992\ln 224 }{ 9 }}e ^{^({\frac{ 1 }{ 9 }\ln \frac{ 224 }{ 222 })2004}}\]

Answer 6

start with
\[ P= A e^{kt} \]
if we start at 1992, we can say
\[ P= 222 e^{k (t-1992)} \]
to find k, use the info
2001, 224 million:
\[ 224 = 222 e^{k (2001-1992)} \]
now we solve for k

Answer 7

in other words:
\[ \frac{224}{222}= e^{9k}\\ \ln \frac{224}{222}= 9 k \\
k= \frac{1}{9} \ln \frac{224}{222}\]

we use that value of k, and t= 2004 to find P

Answer 8

I deleted my previous answer (I solved the rate incorrectly)
I used the web page here:
http://www.1728.org/expgrwth.htm
1992 to 2001 = 9 years
1992 to 2004 = 12 years

First, I solved for the rate

Rate = 10^(log[Ending Amount / Beginning Amount] ÷ time) -1

Ending Amount / Beginning Amount =
224,000,000/222,000,000 =
1.00900900900901

Log(1.00900900900901)=
0.0038950438796

0.0038950438796 / 9 = 0.0004327827

Rate=10^(0.0004327827) -1
Rate = 1.0009970156 -1
Rate = .0009970156


ending amt = begn amt * (1+rate)^time
ending amt = 222,000,000 * (1.0009970156)^12
ending amt = 224,670,663

Answer 9

If I am right, I have the values of A, k, and t as follows:
\[\large \color{#94CA83}{ A=222}\\\Large \color{#94CA83}{ k=9.965×10^{-4}}\\\LARGE \color{#A3B97C}{ t=2004-1992=12}\\\huge \color{#BA9F71}{P=(222)*[e^{(9.965*10^{-4})*(2004-1992)}]}\\\Huge\color{#C2966D}{ P=224.671}\]

I hope I got it!

I am thankful to all of you for giving a helping hand!

Answer 10

We don't really need to force ourselves to explicitly use `e` in problems like these.

In 9 years the population has grown by a factor of 224/222, so the exponential function would be :

\[P(t) = 222 (224/222)^{t/9}\]
Notice that plugging in t = 9 gives you P(9) = 224 as desired.
To get the population in 2004, simply replace `t` by `12`..

Answer 11

So could I have the same answer as I got?

Answer 12

Yes, they are same functions. Mine just looks a lot nicer with all whole numbers :)

Answer 13

Thanks a lot, @ganeshie8!