Question

It is observed that the activity of the radioactive element decreases exponentially according to the formula A(t) = A(10)^-kt, and decreases to 0.35 of its original quantity in 5h. Find the half-life of the element.

Question can be seen here #4: http://tupper.vsb.bc.ca/math/text/p484.pdf

Answer 1

I have a slow and stupid method...
A(t) = A(10)^-kt
A(5) = A(10)^-k(5)
0.35 A = A(10)^-k(5)
0.35 = (10)^-k(5)
log 0.35 = -5k log 10 Note that log10 =1
log 0.35 = -5k
k = -(log 0.35)/5

For half life, A(at half life) = 1/2A
0.5A = A(10)^-kt
0.5 = (10)^-kt
log 0.5 = -kt log 10
log 0.5 = -kt
t = log 0.5 / (-k)
= log 0.5 / -[ -(log 0.35)/5]
= 5log 0.5 / log 0.35
= Can you get it now?

Answer 2

Is that the answer, or is there more?

Answer 3

there is more...

Answer 4

I mean you need to simplify this : 5log 0.5 / log 0.35

Answer 5

I got 3.301260111

Answer 6

same here

Answer 7

THANK YOU SO MUCH!!! I dont want to trouble you any more, thanks for your help Callisto! :)))

Answer 8

Welcome, it's not a trouble though :)

Answer 9

Im prolly not gonna come on OS that much any more because Im done ALL OF MY WORK! YAYY!

Answer 10

Hmm, you can come and help the others :)