a colony of bacteria living on a petri dish under optimal conditions doubles in size every ten minutes. At noon on a certain day, the Petri dish completely covered with bacteria. At what times (to the nearest hour, minute and second) was the percentage of the plate covered by bacteria: 50%, 25%, 5% or 1%

1 answer

amount = c (2^(t/10)), where t is in minutes and c is the initial amount

so when covered by 50% ....

.5 = 1 (2^(t/10)
take log of both sides

log .5 = log 2^(t/10)
log .5 = (t/10)log 2
t/10 = log .5/log2 = -1
t = -10
So it was covered to 50% 10 minutes ago, or at 11:50 am

when covered at 5%
.05 = 1 (2^(t/10))
log .05 = log 2^t/10
t/10 = log .05/log2 = -4.32192..
t = -43.2192 minutes ago

at appr 11:16 am

do the others in the same way