Question

@Will.H

Answer 1

@563blackghost

Answer 2

they give you some numbers. Can you match any of the numbers to the letters in the formula ?

Answer 3

@phi. I would also like to know about this. But can you like solve it and explain your steps. That would be waay faster

Answer 4

A is the initial number amd k is the rate of change and t is time

Answer 5

Go on

Answer 6

whats e ?

Answer 7

e is an important number. It's on your calculator (assume you know it)

you have this equation
\[ P= A e^{kt} \]

they tell you what A is, what P is , and what t is
we need to "plug" those numbers into the equation (as the first step)

any idea on what numbers go with A, P and t ?

Answer 8

p is 6500 and a is 5000

Answer 9

yes, and t ?

Answer 10

"t" stands for time

Answer 11

18 hours?

Answer 12

re-read the info. we want the time when p is 6500 (that is not at t=18)

Answer 13

oh 2 ?

Answer 14

yes.

we put the numbers in to get
\[ 5000 e^{2k} = 6500 \]

the idea is we are solving for the "k" that makes that true
once we know "k", we have a formula that we use for any t
(in this case, we will use the formula to find P when t=18)

to solve for k, we first divide both sides by 5000
what do you get ?

Answer 15

1.3?

Answer 16

you get
\[ e^{2k}= 1.3 \]

(try to always write out the full equation)

next, we have "k" as an exponent. to "get at it" we use ln (natural log)
in other words, we apply ln to both sides:
\[ \ln e^{2k} = \ln 1.3\]

by definition , the ln(e^x) is x
i.e. you get the exponent. You should memorize that.

any way:
\[ 2k = \ln 1.3 \]
can you solve for k ?

Answer 17

btw, you will need a calculator
to find ln(1.3)

Answer 18

k=0.13118?

Answer 19

yes. so the formula is
\[ P= 5000 e^{0.13118\ t} \]
at t=0 we will get P= 5000 (at the start we have 5000)
at t=2 we will get P= 6499.97 (almost 6500. we would need more digits for k to get a closer number. but that is close enough)

any idea what we get for P when t= 18 ?

Answer 20

no?

Answer 21

do you have a calculator ?

Answer 22

oh 6332? is the number supposed to decrease ?

Answer 23

no, the number will "zoom up"
exponential growth is very fast

to be as accurate as possible,
type into google
5000*exp(18*0.5*ln(1.3))=

Answer 24

i got 53022.496865

Answer 25

0.5*ln(1.3) is k (but the calculator will keep as many digits of accuracy as it can)

on your calculator you would do
5000* e^(18*0.131182132)

Answer 26

yes, that looks good.
rounded to the nearest whole number, that is
53022

Answer 27

thats the anwser ?

Answer 28

yes