Question

A wooden artifact from an ancient tomb contains 45 percent of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? (The half-life of carbon-14 is 5730 years.)

Answer 1

The whole half life every \(k\) years, is expressed with: \[
\left(\frac 12\right)^{t/k}
\]

Answer 2

Because then when \(t=k\), you get: \[
\left(\frac{1}{2}\right)^{k/k} = \frac 12
\]

Answer 3

Your amount \(a\) at time \(t\), can be expressed as\[
a(t) = a_0\left(\frac{1}{2}\right)^{t/k}
\]Where \(a_0=a(0)\), the initial amount.

Answer 4

In our case, we want it so that \(a(t) = 45\%\cdot a_0\)

Answer 5

So, putting it all together: \[
0.45a_0 = a_0\left(\frac{1}{2}\right)^{t/5730}
\]And we just solve for \(t\).

Answer 6

i don't know how to solve for t

Answer 7

Do you know what a logarithm is?

Answer 8

this problem is due in 5 min…i just need to know the answer and then ill figure out how i got it

Answer 9

Oh hell no.

Answer 10

Is it homework or a test?

Answer 11

hw

Answer 12

Taking logarithm of both sides gives: \[
\log_2(0.45) = -\frac{t}{5730} \implies t = -5370\log_2(0.45)
\]

Answer 13

so the answer is?

Answer 14

Solve it yourself, please.

Answer 15

(1/2)^(t/5730) = 45
t=??
this also gives same answer
da/dt=-ka, suppose u are losing some factor k
ln a= -kt +c
A=ce^(-kt)
A(0)=100
A(5730)=50
A=100*e^(-ln 2/5730 * t )
45=100*e^(-ln 2/5730 * t )
t=?

Answer 16

I actually gave the answer, I just didn't approximate it. LOL

Answer 17

Though my answer doesn't quite make sense... let me double check it.

Answer 18

Nevermind, I think my answer makes sense now.

Answer 19

ye

Answer 20

e^(-ln 2/5730 * t ) = (e^-ln2)^(t/5730) = (1/2)^(t/5730)