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Mathematics 16 Online
OpenStudy (anonymous):

will give a medal! A country's population in 1992 was 222 million. In 2001 it was 224 million. Estimate the population in 2004 using the exponential growth formula. Round your answer to the nearest million. P = Aekt

OpenStudy (anonymous):

What about the last question??? The way to solve this problem is nearly exactly the same as the last one.

OpenStudy (anonymous):

im on this question

OpenStudy (anonymous):

Help me help you, do you understand what that formula represents, what the goal is of the problem, etc etc. Please tell me what you know and I will tailor my approach to that

OpenStudy (misty1212):

HI!!

OpenStudy (anonymous):

I understand, but if you don't understand the last problem the solutions for the two are almost exactley the same.

OpenStudy (anonymous):

i dont understand the problem the equation

OpenStudy (misty1212):

you can do it without using \[A=Pe^{kt}\] if you like

OpenStudy (anonymous):

Hello @misty1212

OpenStudy (anonymous):

i dont know how to use that either lol

OpenStudy (misty1212):

lets make it easier and call 1992 year zero

OpenStudy (misty1212):

the beginning population is 222 that is your \(P\)

OpenStudy (misty1212):

lol no, not the way they asked the question, that is your A

OpenStudy (anonymous):

ok lol

OpenStudy (misty1212):

\[P=1222e^{kt}\] but you don't know \(k\) that is what you have to find

OpenStudy (misty1212):

typo but whatever

OpenStudy (misty1212):

then since 9 years later it is 224 you can solve \[\large224=222e^{9k}\] for \(k\)

OpenStudy (anonymous):

so divide?

OpenStudy (misty1212):

yeah divide take the log divide by 9

OpenStudy (anonymous):

each sides by 9?

OpenStudy (anonymous):

@ix.ty Do you see why you need to take the log?

OpenStudy (anonymous):

Please make sure to take the log before dividing by 9

OpenStudy (anonymous):

i dont know the log

OpenStudy (misty1212):

\[224=222e^{9k}\\ \frac{112}{111}=e^{9k}\\ \ln(1.009)=9k\\ k=\frac{\ln(1.09)}{9}\]

OpenStudy (misty1212):

oh, if you don't know about logs you can's use this method

OpenStudy (anonymous):

@misty1212 this is the same issue that cam up on the previous problem I was working with @ix.ty

OpenStudy (anonymous):

Ok as I discussed before the natural logarithm is the INVERSE function of the exponential function.

OpenStudy (anonymous):

@ix.ty Do you know what an inverse function means? If so, could you tell me and perhaps give an example please.

OpenStudy (anonymous):

it reverses another funtion f(x)=-1/3x+1 youll but a y infront of the "fx"

OpenStudy (anonymous):

Exactley, an inverse function reverses the action of a function. But to make it a little more rigorous, given an equation of the form y=f(x) where given an value x the function f gives me a value y that is related in a one-to-one fashion with x..... the inverse function (forgoing the discussion of domain and other issues) f^-1(y)=x is how given the y value I can find the corresponding x value that is related to it in a one-to-one fashion

OpenStudy (anonymous):

Ok so if f(x)=x^3 then in order to find the value y=x^3 I have to apply the inverse function which happens to be the cube root: i.e. (y)^(1/3)=(x^3)^(1/3)=x^(3/3)=x

OpenStudy (anonymous):

SO.... the exponential function that is at the cent of this who exponential growth and decay has an inverse function called the logarithm

OpenStudy (anonymous):

Now lets forgo for the moment the idea of e and the natural exponential and the natural logarithm and ask ourselves the question what does exponential growth and decay mean. You first, what does exponential growth mean, and can you give me an example please?

OpenStudy (anonymous):

Sorry the grammar/spelling is so poor, but typing in this text box causes it to bounce around for some reason and given I think faster than I type I sometimes gloss over letters and words sometimes..... what I meant to say above was the exponential function is at the center of of all these exponential growth and decay questions

OpenStudy (anonymous):

The reason why I am trying to go into detail is if you don't understand what it is and where it comes from, these problems will continually give you trouble

OpenStudy (anonymous):

@ix.ty Hello?

OpenStudy (anonymous):

Alright, well I tried twice now to help, but it is a two way street. If I can't get any reciprocity, then I can't help.

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