Question

The population of Guatemala in 2000 was 12.7 million.
1. Assuming exponential growth, what would be the size of the population after time t (measured in years after 2000) if the population was 30 million in 2020? Answer (in millions): P(t)=
2. Assuming exponential growth, what would be the size of the population after time t (measured in years after 2000) if the population was 30 million in 2075? Answer (in millions): P(t)=

Answer 1

@campbell_st @phi

Answer 2

@Hero @bahrom7893 @precal

Answer 3

whats the difference between the old amount and the initial amount

Answer 4

\[A = A_0e^{kt}\]

\[A = \text{new amount}
\\A_0 = \text{initial amount}
\\k=\text{percent growth or decay rate}
\\t=\text{time(usually in years)}\]

Answer 5

so how would i find p(t)

Answer 6

A = P(t)

Answer 7

\[P(t) = P_ie^{kt}\]

Answer 8

what would k be

Answer 9

You have to find k

Answer 10

how

Answer 11

@zepdrix @robtobey @radar @tcarroll010

Answer 12

@muzzammil.raza

Answer 13

yeah?

Answer 14

Just plug in the numbers and solve

Answer 15

12700000e^(20k)?

Answer 16

I wonder if they want P(t) or P(20)

Answer 17

P(t) is just a general formula. P(20) is already given.

Answer 18

they want p(t)

Answer 19

If they want P(t) as it relates to the first question, then you have to find the values of \(P_0\) and \(k\)

Answer 20

Plug those values into the formula

\[P(t) = P_0e^{kt}\]

Answer 21

You find k by inputting all the values, then solving for k

Answer 22

inputting all the values into what

Answer 23

Into the formula bro.

Answer 24

Okay, I'll show you how to really do it.

Answer 25

They give you all the information you need to find k

Answer 26

They already given you \(P_0\)

Answer 27

\(P_0 = 12.7\)

Answer 28

\(P(20) = 30\)

Answer 29

and t = 20

Answer 30

ok so k=30 for the first one

Answer 31

?

Answer 32

i'm lost

Answer 33

Bro, are you making stuff up

Answer 34

no i am trying to find out what u r doing

Answer 35

\(30 = 12.7e^{k(20)}\)

Answer 36

I never mentioned anything about the value of k

Answer 37

Because it is what you have to find

Answer 38

\[\frac{30}{12.7} = e^{20k}\]

Answer 39

12.7 million right

Answer 40

Now take the natural log of both sides:

\[\ln\left(\frac{30}{12.7}\right) = \ln{e^{20k}}\]

Answer 41

You leave off the million. It would be awkward to include it. We already know we're dealing with millions of people

Answer 42

Continue solving for k.

Answer 43

i am getting a negative number for ln(30/12.7)

Answer 44

It's not negative.

Answer 45

Try again

Answer 46

now i am getting .85959595959595...

Answer 47

Good. Now what?

Answer 48

.04298...

Answer 49

That's the value of k

Answer 50

so whats the answer for the question

Answer 51

Now rewrite the general equation of P(t) but only inlude the values for \(P_0\) and k

Answer 52

\[P(t) = 12.7e^{0.043t}\]

Answer 53

so thats the answer

Answer 54

That's your P(t) for the first one. You're on your own for the second one.

Answer 55

please help me with this one i don't know how to do it

Answer 56

You know how to do it. I showed you how to do the first one. The steps for doing the second one is the same as the first.

Answer 57

i still don't know how to do it. i am basically mentaly challenged when it comes to math

Answer 58

Everyone is mentally challenged when it comes to math. But if you keep telling yourself you can't do something, then you won't be able to do it. But imagine what will happen if you keep telling yourself that you CAN do it.

Answer 59

dr.phil please just help me

Answer 60

i alredy tried to do it twice and got it wrong

Answer 61

I'm sorry to hear that.