SOLUTION: if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
Log On
Question 173141: if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
-----------------------
N(1) = 195e^(k*1) = 825
e^k = 825/195
e^k = 4.230769
Take the natural log to find "k".
k = ln(4.230769)= 1.4423838...
--------------------------------
Therefore N(t) 195e^(1.4423838t)
---------------------------------
Then N(10) = 195e^(14.423838) = 358284370 bacteria
======================
Cheers,
Stan H.
(