Aculture of bacteria obeys the law of uninhibited growth. if 500 bacteria are present initially, and there are 800 after 1 hour, how many will be present after 5 hours? round to the nearest whole number.

Since this culture of bacteria obeys the law of uninhibited growth. Given that 500 bacteria are present initially, and there are 800 after 1 hour, a growth rate of 0.470003629246 (k value) has been calculated. 5243 will be present after 5 hours.

The law of uninhibited growth follows below formula:

A = Pe^(rt)

For your problem, we need to first look for the rate, given the initial and final number of bacteria..

A = Pe^(rt)
800 = (500)(e)^r(1)

(800/500) = (e)^r(1)
ln (8/5) = ln (e) ^ r
r = 0.47

substitute the value of r to get the number of bacteria after 5 hours:

A = Pe^(rt)
A = (500)(e)^(0.47)(5)

A = 5,242.78 or 5,243

No because if you do the higher number minus the lower number you will get a positive. if you do it the other way around you will get a negative number