Question

A country's population in 1995 was 228 million. In 2001 it was 230 million. Estimate the population in 2010 using the exponential growth formula. Round your answer to the nearest million

Answer 1

@whpalmer4

Answer 2

What's the exponential growth formula?

Answer 3

P=Ae^kt

Answer 4

Okay, if we take 1995 as t = 0, what value would we use for t for 2001 and 2010, respectively?

Answer 5

2001=6?
and
2010+15?
Yes?

Answer 6

Yes, I think so.

What will we use for our value for \(A\)?

Answer 7

2010=15*

Answer 8

A is Number of the Population at Time Which The Time Is 0?

Answer 9

right. and that is?

Answer 10

228

Answer 11

yes, if we're dealing in millions.

That gives us \[P = 228 e^{kt}\]and we know that at \(t = 6, P = 230\)
Can you solve for the value of \(k\)?

Answer 12

K Is A Positive Constant..Thats What I Put In My Notes.So A Constant Would Be,I Honestly Dont Know?

Answer 13

I tried punching "I Honestly Dont Know" into my calculator to find the population in 2010, but it just gave a funny look and beeped :-)

Answer 14

Do you know about logarithms and exponentials?

Answer 15

A Bit,Its Just That Im Doing A Online Course Of Algebra So I Dont Take It As Math Next Year..But Ive Never Learned These Type Of Equations..

Answer 16

Okay, we'll do a quick demonstration.

A logarithm is the inverse of an exponential. The logarithm (base 10) of \(x\) is the number \(n\) such that \(10^n = x\). That means the logarithm (base 10) of 100 would be 2, for example. Logs have some handy properties, such as:

\[\log (a*b) = \log(a) + \log(b)\]\[\log(a/b) = \log(a)-\log(b)\]\[\log(a^b) = b\log(a)\]

\[e \approx 2.71828 \text{ is the base of the natural logarithm, which is often denoted} \ln\]

The natural logarithm is very handy in many scientific and mathematical contexts for reasons I won't try to cover, you'll just have to trust me :-)

To solve this equation, all you need to know is how to punch the buttons on a scientific calculator or a website that offers similar functionality. Here's how:

\[P = 228 e^{kt}\]\[230 = 228 e^{k*6}\]Divide both sides by 228\]\[\frac{230}{228} = e^{6k}\]Now we'll take the natural logarithm of both sides
\[\ln \frac{230}{228} = 6k * \ln e\](by the last property I listed) and because the logarithm of the base of the logarithm is 1 (x^1 = x, right?) that gives us
\[\ln \frac{230}{228} = 6k\]\[\frac{1}{6} \ln\frac{230}{228} = k\]That's just a simple bit of button-pushing on the calculator, and the result is approximately 0.00145561. Our formula becomes \[P = 228 e^{0.00145561t}\] and now we just plug in \(t=15\) to get our answer for the estimated population in 2010.

Answer 17

If you have a scientific calculator, it probably has an e^x and ln x button pair. If not, you could use http://www.wolframalpha.com which will allow you to type it in like this:

http://www.wolframalpha.com/input/?i=228+e%5E%280.00145561t%29%2C+t+%3D+15

Answer 18

Woah.
I Got 233..After I Round?correct?

Answer 19

Yes, that's what I get as well. You should try the formula at t = 6 to check our work on finding \(k\). If we did it right, you should get 230.

Answer 20

Yes Sir..Works Out Correct When I Plugged In and Checked..Thanks You So Much..and You Sir Have A Wonderful Day.. :)