HELP...Modeling with exponential and logarithmic functions?

Question

(Using the Radioactive decay Model)

Radioactive radon After 3 days a sample of radon-222 has decayed to 58% of its original amount.

a) What is the half life of radon-222?

b) How long will it take the sample to decay to 20% of its original amount?

Answer 1

Hi,

a) 58 = 100(½)^x will show what this exponent would be to solve this problem.
.58 = (½)^x
log .58 = log (½)^x
log .58 = x log (½)
log .58/log .5 = x
.7858751946 = x
That means the exponent must equal .7858751946 when the number of days is divided by the half life. So if 3 days were used to get to 58% remaining, then 3/half life = .7858751946. Solving this for half life gives

3/half life = .7858751946
3 = .7858751946 *half life
3/(.7858751946) = half life
3.8174 = half life

The half life of radon-222 is 3.8174 days.

b) 20 = 100(½)^(x/3.8174)

.2 = (½)^(x/3.8174)

log .2 = log (½)^(x/3.8174)

log .2 = x/3.8174 * log (½)


3.8174 log (.2)
--------------------- = x
log (½)

8.8637 = x

It would take 8.8637 days for the sample to decay to 20% of its original value.

I hope that helps!! :-)

Answer 2

a)
.58 = (.5)^(3/h)
log(.58) = (3/h)log(.5)
log(.58)/log(.5) = 3/h
h = 3log(.5)/log(.58) = 3.817 days.

b) let r be the daily decay rate, so
.58 = r³
.20 = r^t
log(.58) = 3log(r)
log(.20) = tlog(r)
log(.58)/log(.20) = 3/t
t = 3log(.20)/log(.58) = 8.8637 days.

Answer 3

a. 3.824

b. 2.2 half-lives or 8.4128