Find the half-life (in hours) of a radioactive substance that is reduced by 30 percent in 80 hours

Answer 1

amount = a(1/2)^(t/k), where a is the starting amount, and k is the half-life period

now, when t=0, amount = 1
1 = a(1/2)^0 , so a = 1
after 80 hours:
.7 = (.5)^(80/k)
log .7 = (80/k)log .5
80/k = log.7/log.5
k/80 = log.5/log.7
k = 80log.5/log.7 = about 155.5 hours