Question: The # of bacteria grows exponentially @ 1% per hr. Assuming 10,000 bacteria are present initially...

The # of bacteria grows exponentially @ 1% per hr. Assuming 10,000 bacteria are present initially, find:





a) the # of bacteria present at any time T.


b) the # of bacteria present after 5 hrs


c) the time req. for # bacteria to reach 45,000?





Please help, and please explain! This is the first question in my homework set and I just want to make sure I'm doing it the right way.

Question: The # of bacteria grows exponentially @ 1% per hr. Assuming 10,000 bacteria are present initially...
a) # of bact = 10000*(1.01)^T





as it is an exponenetial growth you can put it straight into this equation.





b) # of bact at 5 hours = 10000*(1.01)^5 = 10,510





c) 10000*(1.01)^T =45000 =%26gt; (1.01)^T =4.5 =%26gt;





T*log(1.01) = log 4.5 =%26gt; T = log4.5/log1.01 =%26gt;





T = .6532/.0043 =%26gt; T = 151.2 Hours.
Reply:total= 10000*(1.1)^t,








where t is hours.





so, a) is y=10,000(1.1)^x





and you can solve the rest from there.
Reply:I retract the following answer. My answer would be for CONTINUOUS exponential growth. I didn't pay close enough attention, but your problem says 1% per hour, which is growth compounded once per hour. The other answers seem correct. But maybe this will help with another problem on your homework:





a) N(t) = 10000e^[(.01)t]





b) Just plug 5 in for t and solve in your calculator.





c) 45000 = 10000e^[.01t]





4.5 = e^[.01t]





ln (4.5) = .01t





t = (ln [4.5])/(.01)





Plug into a calculator ( I don't have one handy). TIme is in hours! Hope this helps.
Reply:a) N = 10,000(1.01)^T


b) N(5) = 10,000(1.01^5) = 10,510


c)


45,000 = 10,000(1.01)^T


(1.01)^T = 4.5


Tln(1.01) = ln(4.5)


T = 151.1585312 hr or 151 hr 9 min 31 sec