Question

okay so this is the last problem i need to do for my homework and i have no idea what to do for it....im guna take a screenshot and post it and hopefully someone can help:D

Answer 1

half life of drug is \(6.8\) hours
initial dose is \(55\) this can be modelled as
\[\large A(t)=55\times \left(\frac{1}{2}\right)^{\frac{t}{6.8}}\]

Answer 2

there are other ways to do it involving \(e\) but this is simplest, since i only used the numbers given, not solving exponentials or anything complicated

Answer 3

i vote the simple way! lol

Answer 4

so for A put \(t=24\) and compute

Answer 5

is A(t) the same as An?

Answer 6

i am confused about \(A_n\) because that says "after the nth dose"
how can you do that without knowing how often the drug is administered?
i have to say also that i am not sure how to do it if we knew that, but at least we could start

Answer 7

yeah i dont understand why this question is in my homework ive never seen any of this in class

Answer 8

looks like a geometric sequence problem see attached

Answer 9

i dont think ive seen this stuff before...but proceed lol....

Answer 10

proceed? i have no idea how to do this, but at least we can start

Answer 11

Yeah sorry idk how to do this either... but i have seen it before...

Answer 12

\(C_0=55\) for one
unfortunately we need \(r=e^{-kL}\) and i don't know exactly what \(L\) is but we can find \(k\) for sure

Answer 13

proceed explaining was what i meant...oh i thought u knew lol but yeah lets start!

Answer 14

oh maybe \(L\) is just \(t\)

Answer 15

to find \(k\) or as they call it \(-k\) since you know the half life is \(6.8\) you can solve
\[.5=e^{6.8k}\] for \(k\)

Answer 16

you get the answer more or less in one step
\[k=\frac{\ln(.5)}{6.8}\]

Answer 17

or \(-k=-.1002\) rounded

Answer 18

so what exactly are we finding right now?

Answer 19

\(-k\) for the formula
\[C_0e^{-kt}\]

Answer 20

and what does that formula tell me? im sorry im just trying to understand this question lol

Answer 21

i am just following along with what is written in the pdf i sent
looks like the concentration immediately after the nth dose is
\[C_0\sum_{i=1}^{n-1}e^{-ikt}\]

Answer 22

we know \(C_0=55\) and we just found \(-k=-.1002\)

Answer 23

and it also says the concentration immediately before taking the nth dose is
\[C_0\sum_{i=0}^{n-1}e^{-ikt}\]

Answer 24

no i got those backwards!!

Answer 25

after then nth dose start at \(i=0\)\[C_0\sum_{i=0}^{n-1}e^{-ikt}\]

Answer 26

before the nth dose start at \(i=1\)
\[C_0\sum_{i=1}^{n-1}e^{-ikt}\]

Answer 27

so before is
\[A_n=55\sum_{i=1}^{n-1}e^{-.1002it}\]

Answer 28

@agent0smith how am i doing? first time out with this

Answer 29

oh i think bad, because i think you are supposed to actually add this up using the previous formulas

Answer 30

ok we can do that
lets do the last one first, it is easiest
last one is the limit of the sum, which is
\[\frac{55}{1-e^{.1002}}\]

Answer 31

damn i forgot the minus sign
last one is
\[\frac{55}{1-e^{-.1002}}\]

Answer 32

im a bit confused im not guna lie:( im sorry ur working so hard!

Answer 33

I'm not experienced enough with this to know... i think i've seen it a small handful of times, don't think i actually solved one myself though...

Answer 34

yeah i am too, but if you read the pdf most of it comes from finding the sum and the limit of the sum
the real work was finding \(-k\) in the model
\[A(t)=55c^{-kt}\] which i avoided doing at the beginning

Answer 35

i am still a little confused by this \(L\)

Answer 36

before then nth dose when you sum up you get
\[55\frac{e^{-1.002L}-e^{-.1102nL}}{1-e^{-.1102L}}\]

Answer 37

after the nth dose it is \[55\frac{1-e^{-.1102nL}}{1-e^{-.1102L}}\]

Answer 38

oooh it says you take it every day, and the time is measured in hours! i should learn to read

Answer 39

replace all \(L\) by \(24\) then

Answer 40

okay let me try plugging that in and see if its right

Answer 41

before the nth dose
\[55\frac{e^{-1.002\times 24}-e^{-.1102\times 24}}{1-e^{-.1102\times 24}}\]

Answer 42

ooops should be an N in there

Answer 43

im not able to put an n in my answer:/

Answer 44

\[\large 55\frac{e^{-1.002\times 24}-e^{-.1102\times 24n}}{1-e^{-.1102\times 24}}\]

Answer 45

for B it asks for \(A_n\) so you have to write in terms of \(n\)

Answer 46

\[A_n=55\frac{1-e^{-2.4048n}}{1-e^{-2.4048}}\]

Answer 47

\(P_n\) is similar but instead of 1 you put \(e^{-2.4048}\) up top

Answer 48

its not right:(

Answer 49

actually hold on

Answer 50

try this for the last one
\[\frac{55}{1-e^{-2.4048}}\]

Answer 51

or \(60.4584\)

Answer 52

nope neither are right:(((

Answer 53

sorry i am not better at this, but i am learning as i am reading
maybe i can find a better source, with a worked out problem

Answer 54

omgsh dont be sorry!!! ur guess is better than mine!

Answer 55

Good job @satellite73

Answer 56

ok i think i have an easier way
lets answer the first question first

Answer 57

okay:D

Answer 58

\[\large (\frac{1}{2})^{\frac{24}{6.8}}=.0866\]

Answer 59

that should answer question one
try it and lets see

Answer 60

nope:(

Answer 61

goddamn it , it has to be right

Answer 62

im sorry:(

Answer 63

half life is \(6.8\)
24 hours later where will be \(\left(\frac{1}{2}\right)^{\frac{24}{6.8}}\) amount left

Answer 64

of that much, i am certain

Answer 65

can you try \(.09\) or \(.087\) maybe?

Answer 66

nope neither worked:(

Answer 67

are you supposed to enter is as a fraction? does that make a difference?
\(\frac{9}{100}\) or \(\frac{87}{1000}\)

Answer 68

im not sure let me try!

Answer 69

nope:(

Answer 70

then i am stuck
but i am 100% sure
let me read the question again
half life is 6.8
24 hours
after 6.8 hours half is left
after 13.6 hours 1/4 is left
after 20.4hours 1/8 is left
after 24 hours \(( \frac{1}{2})^{\frac{24}{6.8}}\) is left

Answer 71

if we can't find that number, then we cannot continue
sorry

Answer 72

omg that was right!!! we just kept simplifying it!

Answer 73

??

Answer 74

i swear i put that in like 5 tries ago and it didnt work!

Answer 75

im so sorry!

Answer 76

what worked exactly? we will work with that number

Answer 77

(1/2)^(24/6.8)

Answer 78

christ on crutch

Answer 79

oh i know what i did! i put in the simplified decimal point you gave me instead of that!

Answer 80

ok do you think that means we have to continue using that? or can we use a decimal approximation?

Answer 81

if the decimal approx. doesnt work then we can just enter (1/2)^(24/6.8)

Answer 82

ok
after the nth tables it should be
\[55\frac{1-.0866^n}{1-.0866}\] if we get to use the decimal
this is AFTER

Answer 83

nope it didnt

Answer 84

work

Answer 85

we could try it with the power
or try it with n - 1 instead of n, but i am pretty sure than n is right

Answer 86

for that matter we could divide \(55\) by \( 1-.0866\)

Answer 87

or put \(.9134\) in the denominator
hard to know exactly what this wants

Answer 88

lets try the last one
i think it is \(\frac{55}{1-.0866}=60.21\) which is pretty close to what we had the other way

Answer 89

hold on ill brb

Answer 90

last one isnt right:(

Answer 91

ok then i give up

Answer 92

seriously thank you so much for trying and all of your hard work! i am very grateful and appreciative!

Answer 93

want to try
\[\frac{55}{1-(\frac{1}{2})^{\frac{24}{6.8}}}\]

Answer 94

yw
i really don't know why that one didn't work
we got essentially the same answer doing it two different ways

Answer 95

or add more decimals?
http://www.wolframalpha.com/input/?i=55%2F%281-%281%2F2%29^%2824%2F6.8%29%29

Answer 96

nope for ((55)/(1-(1/2)^(24/6.8)))