Exponential Growth and Decay?

Question

Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 78% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment.

No idea where to start on this one?

Answer 1

let t be number of years the former vegetation is dead, r the residual radioactivity as a decimal fraction. halflife equation is
r = (1/2)^(t/5730)

note that when t = 5730, r = 1/2. so solve
0.78 = (0.5)^(t/5730)
log 0.78 = (t/5730)log 0.5
t/5730 = log(0.78)/log(0.5)
note that the ratio of the decay times is equal to the ratio of the logs of the residual radioactivity
t = 5730 log(0.78)/log(0.5)
t = 2053.9 years

Answer 2

definite, the exponent is the "t" being superior by potential of the a million/5th. A(t) represents the quantity after time t. So 4 is the preliminary quantity through fact A(0) = 4. a million/2 potential to shrink by potential of a million/2 (with exponent t/5). i think of you're engaged on some questions approximately radioactive decay? If definite, then the a million/2-life is 5, considering the fact that while t = 5, A(t) = A(0) / 2, i.e., turns right into a million/2 of the preliminary quantity. the two t and the "5" ought to have a unit, le us say in 300 and sixty 5 days. So after 2 many years, the entire quantity left is A(20) = 4 * (a million/2) ^(20/5) = 4 * (a million/2) ^ 4 = a million/4.

Answer 3

Start by using the half-life information to find your rate of decay, k.
If Ao is your initial amount, then in 5730 years you will have 1/2Ao.
1/2Ao = Ao e^(5730k) divide out the Ao
1/2 = e^(5730k) take the ln of both sides
ln (1/2) = 5730 k
-.00012 = k

Now you have that A = Aoe^(-.00012t)
If you have 78% of the amount left, then A= .78 Ao
.78Ao = Aoe^(-.00012t)
.78 = e^(-.00012t)
ln.78 = -.00012t
2070.5 yrs = t

Answer 4

The general formula for exponential decay is
N = A(.5)^(t/h) where t = time elapsed, h = half-life,
A = starting amount and N = final amount.
.78A = A(.5)^(t/5730)
.78 = (.5)^(t/5730)
log(.78) = (t/5730)log(.5)
5730(log(.78)/log(.5)) = t
t = 2053.941253. Approximately 2050 years.

Answer 5

current mass = (original amount)(1/2)^(number of years/5730)
So,
if you say that 78 percent of it mass is conserved=
current mass/original amount = .78

so,

.78 = (1/2)^(n/5730)
solve for n