A(t) = ke^-0.001t, where k is the constant.?

Question

When a certain radioactive elemnt decays, the amount, in milligrams, that remain after "t" years can be approximated by the function "A" above. Apporximately how many years would it take for an initial amount of 800 miligrams of this element to decay to 400 miligrams.

The answer is 693. I need to know how to get that answer.

BTW, Where are you guys getting .5? Why are you dividing 400 by 800?

The first one to explain how/why to get .5 by dividing 400/800 gets the 10 points.....

Answer 1

800 = 400 e ^ -0.001t
0.5 = e ^ -0.001t
ln 0.5 = -0.001t
t = (ln 0.5)/ -0.001

Answer 2

First answer, Kate G, told you where 0.5 comes from. k is the initial amount, because if you put t = 0 (initial), you get
A(0) = k*e^0
which is k because e^0 = 1.

So Kate G put 800 for k, giving her first line,
then had to divide by 800 to find
e^-0.001t

Give Kate G the 10 points.

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Answer 3

A(t) = ke^(-t*10^(-3))

A(0)=800*10^(-3) g=k

Calculation of th (h=halving time);
400*10^(-3)=800*10^(-3)*e^(-th*10^(-3))=
400/800=1/2=e^(-th*10^(-3))=>
-ln(2)=-th*10^(-3)=>
th=ln(2)/10^(-3)second
=================
because ln(1)=0, ln(e^x)=x

Answer 4

You ask for the half time.
A(t) = ke^-0.001t
A(0) = k * 1 = k = 800 mg
A(t(wished)) = 800 mg * e^-0.001t = 400 mg (given).
So e^-0.001t = 0.5. Take the natural logarithm:
-t/(1000y) = -0.693 so
t = 693 years

Th

Answer 5

1/2 = ke^(-.020) and solve for k (about .5 without a calculator ☺)


Doug

Answer 6

because ln(0.5)*-1000 = 693.14718 ..

I would say the answer is 694 years