Question 198168
A bacteria population starts at 2032 and decrease by 15% everyday.
Write a function representing the number of bacteria present each day.
f(t) = 2032(1-.15)^t
f(t) = 2032*.85^t
:
After how many days will there be fewer than 321 bacteria?
2032*.85^t = 321
.85^t = {{{321/2032}}}
.85t = .1578
t*log(.85) = log(.1578)
-.07058t = -.80189
t  {{{(-.80189)/(-.07058)}}}
t = 11.36 days
We can say after 12 days there will be less than 321
:
Check solution on calc: enter 2032*.85^12 = 289 left which is less than 321
in 11 days: 2032*.85^11 = 340, more than 321