Question

help pweeeeeeeeeeeeeease

In 1994, 24 samples from the archeological site at Oslonki, Poland were dated using carbon 14. The archeological site is significant for the large quantities of copper implements and jewelry found there. One of the samples had 47.8% of the carbon 14 that would normally be present in a living organism. Approximately, how old is the sample? (Carbon 14 decays approximately 1.202% every 100 years.)

Answer 1

@nincompoop

Answer 2

@satellite73

Answer 3

1.202%?

Answer 4

8319.47

Answer 5

=(

Answer 6

im wondering if theres a way to use logs for this

Answer 7

One second let me look up the formula for compound decay

Answer 8

kk thank u

Answer 9

Use this

Answer 10

wait so are we using the slope formula for this

Answer 11

ok

Answer 12

so we have y = m_0e^(rx)

where
x = time
y = amount at time t
m_0 = initial amount of carbon 14 which would be 100%
r = rate of decay

we know
m_0 = 100%
r = ?

we know point (100, 1.202%)
so we can solve for r

Answer 13

once you solve for r you can simply sub the percentage into y and solve for x to find the time it took to decay

Answer 14

follow?

Answer 15

hmmmmm well the thing is i never used that formula u gave me so im tryna let it sink in

Answer 16

we used a different method in class we used excel =/

Answer 17

so you never saw it in class? because this would exponential decay since it is radioactive

Answer 18

no we used excel

Answer 19

Well this formula is pretty common in grade 12 functions so I dont know and besides the algebra isn't hard it cant hurt to learn it now

Answer 20

ok gotcha

Answer 21

Make an attempt at solving for r and I will tell you if you are on the right track, just use logarithms to solve

Answer 22

yes we used logs im trying to refresh my memory how would i go about using logs that is way more simpler

Answer 23

first divide both sides by 100 then take the natural logarithm of both sides

Answer 24

natural logarithm being

ln(x)

Answer 25

log100= log 1.01202

Answer 26

remember the rule

ln(e^x) = x

Answer 27

log100=/= log 1.01202

Answer 28

so its 385.42

Answer 29

the answer is 6100 years though i dont know where i messed up

Answer 30

sorry I made a mistake the point we have is (100, 98.798%)

Answer 31

where did u get that number from

Answer 32

We also know that at 100 years we have 100%-1.202% =98.798% sample left so thus we have another point (100, 98.798%)

Answer 33

oh ok and im guessing since its decay your subtracting?

Answer 34

after the first initial 100 years we have 1.202% of the sample gone thus we are left with 98.798% of the sample left

ok I got it

Answer 35

It's looking for an approximation and gives the decay rate with a precision of ±100 years.
I think

\(\large 0.478=1.000(1-0.01202)^t\)

with t in units of hundreds of years ought to do it.

Answer 36

so you know at

t = 0
y = 100%

so we have that

m_0 = 100%

so we have the equation is

\[y = 100e^{-rt}\]

where 100 = 100%

now we sub in the point we have into the equation to solve for the rate or r

since we know the point

y = 100% - 1.202% = 98.798%
x = 100 years

we can sub that point into the equation to solve for r

\[98.798 = 100e^{-100r}\]

You should get
r = 0.000120928

Then we have the final formula

\[y = 100e^{-0.000120928(t)}\]

with this formula/function we can simply sub in the % carbon 14 of the sample to find its date

Answer 37

sorry for taking so long to answer this but wolframalpha was acting really strange

Answer 38

its fine im just reviewing everything u wrote down

Answer 39

Try \(\large t=\frac{log(0.478)}{log(0.98798)}\)

Answer 40

oh ok i guess u cant use logs for this then

Answer 41

oh nvm i guess we can

Answer 42

61.04 is what i got

Answer 43

sorry that is incorrect yomamabf

Answer 44

t is in units of 100 years . . .

Answer 45

seriously follow my method, not to be a jerk to cliff sledge but I got a plausible answer and this equation is for radioactive decay

Answer 46

your answer is correct than cliffsledge

Answer 47

You're not being a jerk at all, @Australopithecus , I just think you are working to hard on a question that asks for an approximation with a margin of error of ±100 years.

Answer 48

your answer is correct then cliffsledge

Answer 49

you are probably right cliffsledge ha

Answer 50

but it really isn't that hard to use this formula, my laziness just got the better of me

Answer 51

I think what you're doing is probably correct too. Here is a summary of an alternate method:

Generic exponential model, \(\large A=P(1+r)^t\)
A=0.478, the 47.8% remaining
P=1.000, the 100% started with
r=-0.01202, the 1.202% decay rate
t= time (in 100s of years).

Plug in the knowns and solve for t.

Answer 52

yeaaaaaaaaaaaa thats what we used in class

Answer 53

Since it's approximate to ±100 years, a continuous compounding model isn't appropriate.

Answer 54

If a half-life was given, then using e and natural logs would work nicely.

Answer 55

the answer however is 6100 years and its not adding up

Answer 56

Why not? I got 6104 years when I did it.

Answer 57

hang on lemme try this again

Answer 58

sorry i plugged everything in and i got a .47232

Answer 59

? Hmm..

\(\large A=P(1+r)^t\)

\(\large 0.478=(1-0.01202)^t\)

\(\large log_{0.98798}(0.478)=log_{0.98798}((1-0.01202)^t)\)

\(\large t= log_{0.98798}(0.478) = log(0.478) \div log(0.98798)\)

Answer 60

oh u used logs? ok ill try it again

Answer 61

Have to. Need to solve for the exponent, that's what logs do.

Answer 62

ohhhh i finally got it thank u so much <333 id give u a medal but nincompoop stole it and deleted his replies

Answer 63

You're welcome.