Question

A country's population in 1990 was 123 million.In 2002 it was 128 million.Estimate the population in 2013 using the exponential growth formula.(P=Ae^kt).Round to the nearest millionth..

Answer 1

@whpalmer4 You Helped Me Yesterday,Can You Help Me Today Please.

Answer 2

Indeed, exact same approach as yesterday...

Answer 3

Give it a try, I'll help if you get stuck.

Answer 4

Im Having A Bit Of Difficulty Pluging In The Numbers For Variables. @whpalmer4 Id Be Very Great full If You Could Walk Me Through It.

Answer 5

What is the initial value? Assign that to A

The initial year is 1990. 2002 is the year for which we have another's data point. We can write \[P=Ae^{k(t-1990)}\] which is a bit fancier than yesterday's equation. In this one we plug in the year directly...

Answer 6

A=123?

Answer 7

In 2002, the population in millions is 128, so we can solve for \(k\)
\[128=123e^{k(2002-1990) }\]
Did any of the quick logarithm tutorial stick well enough to solve that?

Answer 8

No ;( I Remember That A Logarithm Is The Inverse of an Exponent..and x=10

Answer 9

Okay, if we divide both sides by 123, we get
\[\frac{128}{123}=e^{12k}\]
Take the natural log of both sides
\[\ln(\frac{128}{123})=12k\]

Answer 10

Evaluate the left side with your calculator and divide it by 12 to get \(k\)

Answer 11

Okay.Im Following.

Answer 12

Now plug the newly found value of k into the formula, set t=2013 and find the answer!

Answer 13

What did you get for your answer?

Answer 14

My Computer Just Restarted But I Got
133

Answer 15

I got 132.762, so rounded to the nearest million that would be 133, just like you. Good job!

Answer 16

Thanks You So Much!Again :)