i have this question and im not sure how to do it i keep getting stuck with these types of questions and i would appreciate some help

Question

(a) The rate of decay of a radioactive substance, S, is proportional to the amount remaining, x. Given that x = a at time t = 0, show that, if the time take for the amount of S to become 1/2 a is T, then at

ln(a/x)= (t/T)ln2

If T= 15 days, calculate the time taken for there to be 10% of S remaining.

ive calculated 24.14 days for the second part im not sure if this is right and i don't know how to do the first part

Answer 1

From the description T is the half-life of the substance
your equation becomes
x = a(2)^(-t/T)
x/a = 2(-t/T)
take ln of both sides
ln(x/a) = ln[2^(-t/T)
ln(x/a) = (-t/T)ln2

then for the second part:
.1a = a(2^(-t/15)
ln .1 = -t/15ln2
t = 49.8

check:
after 15 days .5 is left
after 30 days .25 is left
after 45 days .125 is left

my answer sounds reasonable.

Answer 2

can you explain how you got
x = a(2)^(-t/T)
from the information in the question
and the original equation

ln(a/x)= (t/T)ln2

not

ln(x/a) = (-t/T)ln2

which is what you had??

Answer 3

thanks

Answer 4

Didn't check for any replies til now.

in general, if you have have a doubling or halving situation, you can use a base 2, such as our equation

x = a(2)^(t/k)

if the function is increasing (doubling) then the exponent must be positive,
if the function is decreasing (like our halving) then the exponent must be negative.
Some texts and authors will change the base to (1/2) for half-life or decreasing functions, that way the exponent is always positive.
I used the first option.

Notice that using my equation the value of t comes out correctly as a positive number.