Question

The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.

Answer 1

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: \[\frac{dP}{dt}\alpha P\]
\[\Rightarrow \frac{dP}{dt} = aP\]
\[ \Rightarrow \frac{dP}{P} = adt\]
\[ \Rightarrow \log \left| P \right| = at + C . . . . . \left( 1 \right)\]
Now, 
\[P = N\text{ at }t = 0\]
\[\text{ Putting }P = N\text{ and }t = 0\text{ in }\left( 1 \right), \text{ we get }\]
\[\log \left| N \right| = C\]
\[\text{ Putting }C = \log \left| N \right|\text{ in }\left( 1 \right),\text{ we get }\]
\[\log \left| P \right| = \text{ at }+ \log \left| N \right|\]
\[ \Rightarrow \log \left| \frac{P}{N} \right| =\text{ at }. . . . . \left( 2 \right)\]
According to the question, 
\[\log \left| \frac{2N}{N} \right| = 6a\]
\[ \Rightarrow a = \frac{1}{6}\log \left| 2 \right|\]
\[\text{ Putting }a = \frac{1}{6}\log \left| 2 \right|\text{ in }\left( 2 \right),\text{ we get }\]
\[\log \left| \frac{P}{N} \right| = \frac{t}{6}\log \left| 2 \right| . . . . . \left( 3 \right)\]
\[\text{ Putting }t = 18 \text{ in }\left( 3 \right)\text{ to find the bacteria after 18 hours, we get }\]
\[\log \left| \frac{P}{N} \right| = \frac{18}{6} \log \left| 2 \right|\]
\[ \Rightarrow \log \left| \frac{P}{N} \right| = 3\log \left| 2 \right|\]
\[ \Rightarrow \log \left| \frac{P}{N} \right| = \log \left| 8 \right|\]
\[ \Rightarrow \frac{P}{N} = 8\]
\[ \Rightarrow P = 8N\]