Question

In 360 days the radioactivity of a sample decreases by 74 percent. what is the half life? In 360 days the radioactivity of a sample decreases by 74 percent. what is the half life? @Mathematics

Answer 1

solve
\[.74^t=\frac{1}{2}\] for t in one step

Answer 2

im in calculus 2 not physics

Answer 3

OH using y=Ce^kt

Answer 4

you get
\[t=\frac{\ln(\frac{1}{2})}{\ln(.74)}\]

Answer 5

of course your answer will be in years, not days

Answer 6

840?

Answer 7

hold on i have to read carefully

Answer 8

it says "decreases by 74%" not "has 74% of its initial value" so i was incorrect.

Answer 9

ok

Answer 10

if it decreases by 74% it has only 26% of its initial value. so you have to solve
\[.26^t=.5\] instead

Answer 11

same idea though, get
\[t=\frac{\ln(.5)}{\ln(.26)}\] and again your units are years. but if you want days that is easy enough

Answer 12

i get 187.8 days rounded.

Answer 13

it says that is wrong, but it wont give me the correct answer

Answer 14

how many decimal places does it want?

Answer 15

http://www.wolframalpha.com/input/?i=log%28.5%29%2Flog%28.26%29

Answer 16

its not specific and i already tried all the decimals

Answer 17

I GOT IT

Answer 18

ok maybe try this one
but i don't think it is right
http://www.wolframalpha.com/input/?i=log%28.5%29%2Flog%28.74%29*365

Answer 19

oh what was the problem?

Answer 20

u multiply by the 360 not 365

Answer 21

lordamercy. nothing like the ability to read. i guess i just assumed it was 365 when i read it.

Answer 22

can you show me how you derived .26^t=.5 from the y=Ce^kt formula

Answer 23

yes

Answer 24

first off the C is unimportant, because you want the half life. so it doesn't matter what you start with, at the end you will have half of the original amount. in other words if i start with
\[C\times .26^t=\frac{1}{2}C\] the first step will be to rewrite as
\[.26^t=\frac{1}{2}\]

Answer 25

as for the .26 i reason that if you lose 74% you retain 26% (not much to that, 100 - 74 = 26)

Answer 26

ok so now the next part of the question asks How long will it take for a sample of 100mg to decay to 81 mg

Answer 27

and therefore i don't need the
\[Ce^kt\] formula which requires me to solve for k using logs etc. i just say that every year it is 26% of the previous year, so i use
\[.26^t\] instead of
\[e^{kt}\]

Answer 28

ok now we have to be careful

Answer 29

since i messed up with the year business. we need to use
\[100\times .26^{\frac{t}{360}}=81\]

Answer 30

ya that makes sense not because C is the initial

Answer 31

divide by 100 get
\[.26^{\frac{t}{360}}=.81\]

Answer 32

now*

Answer 33

right, but i have to use an exponent of
\[\frac{t}{360}\] because it decrease to 26% of its current amount every 26 days

Answer 34

360 days rather

Answer 35

solve as before
\[\frac{t}{360}=\frac{\ln(.81)}{\ln(.26)}\]
\[t=360\times \frac{\ln(.81)}{\ln(.26)}\]

Answer 36

ya i got that, let me try to put in the answer

Answer 37

http://www.wolframalpha.com/input/?i=360+log%28.81%29%2Flog%28.26%29

Answer 38

yup got it, thanx alot

Answer 39

yw

Answer 40

i got a couple more question concerning decay and growth, so if you can just keep watch for my posts, i really appreciate your help

Answer 41

no problem, but post quickly because i have to go soon enough

Answer 42

alright

Answer 43

btw you can use
\[Ce^{kt}\] if you like, but it requires finding k, and then when you use the decimal there will be rounding error

Answer 44

you have 700 dollars in your bank account. Suppose your money is compounded every month at a rate of 0.5 percent per month.
(a) How much do you have after t years.

Answer 45

C=700 but how do i set up the rest of the equation