Question

A particular radioactive isotope decays from 180 milligrams (mg) to 156.6 mg in 10 days. Find the half life of the isotope.

Answer 1

What is the answer in days?

Answer 2

The formula I am given is:

\[y=y_{0}(1-r)^{x}\]

Answer 3

oh ok its same process , i just assumed it was exponential
you can use either form really

anyway, instead of solving for "k" first , you would solve for "r"

Answer 4

Ok, so: \[156.6=180(1-r)^{10}\]
?

Answer 5

yes
then
\[\ln (156.6/180) = \ln (1-r)^{10}\]

Answer 6

r=-0.139

Answer 7

hmm r cant be negative because 1-r must be less than one to make it a decay function

Answer 8

the exponent "10" can come out front so
\[\ln (1-r) = \frac{\ln(156.6/180)}{10}\]
then to cancel the log on left side take "e" raised to both sides
\[\large 1-r = e^{\frac{\ln(156.6/180)}{10}} = (\frac{156.6}{180})^{1/10}\]

Answer 9

I get -0.013926207, but how do I calculate the days?

Answer 10

not sure how you are getting a negative number?

but to get the days, you need to solve this equation for "x"
\[(1-r)^x = \frac{1}{2}\]

Answer 11

\[x = \frac{\ln (1/2)}{\ln (1-r)}\]

Answer 12

ok so that is 10 days, but what is the half life?

Answer 13

what is 10 days?

that is the half-life .... its the "x" value when y = 90 or y/y_0 = 1/2

Answer 14

ok originally we found that
\[\ln(1-r) = \frac{\ln (156.6/180)}{10}\]
plug that into
\[x = \frac{\ln(1/2)}{\ln(1-r)}\]
and you get your answer

Answer 15

49.42547845

Answer 16

close enough ... i get 49.77

be careful with rounding...try to store answers or put entire expression in calculator

Answer 17

ok yes I finally got 49.77 also...thank you for your patience...My webwork says it is correct.

Answer 18

yw