The population of the bacteria colony starts at 100 and grows by \[30\% \] per hour.
A.Find the formula for the number of bacteria, P, after t hours.
B.What is the doubling time of this population; that is, how long does it take the population to double in size?
Answer
550.5k+ views
Hint: To solve this question, we need to use the concept of exponential growth. The exponential growth occurs when the growth rate of the value of a mathematical function is proportional to the function’s current value, resulting in its growth with time being an exponential function. We can also say that when the growth of a function increases rapidly in relation to the increase in the total number, then it is exponential.
Formula used:
Systems that experiences exponential growth increase according to the mathematical model:
\[P = {P_0}{e^{kt}}\]
Where, $ {P_0} $ represents the initial state of the system and $ k $ is a constant, called the growth constant.
Complete step by step solution:
We are given that the bacteria colony starts at 100 and grows by \[30\% \] per hour.
Therefore, $ {P_0} = 100 $ and $ k = 0.3 $ .
By using the formula \[P = {P_0}{e^{kt}}\] , we can say that \[P = 100{e^{0.3t}}\] .
Thus, the answer for our first question is: the formula for the number of bacteria, P, after t hours is: \[P = 100{e^{0.3t}}\] .
We also know that the continuous growth rate is nothing but the growth constant.
Therefore, the answer to the third question is: the continuous growth rate for the colony is $ 0.3 $ .
Now, for the second question, we have to find the doubling time.
Therefore, \[P = 2{P_0} = 200\] . Putting this value into the main equation, we will get
\[
200 = 100{e^{0.3t}} \\
\Rightarrow 2 = {e^{0.3t}} \;
\]
Now, we will take logarithms on both the sides.
\[
\Rightarrow \ln 2 = \ln {e^{0.3t}} \\
\Rightarrow \ln 2 = 0.3t\ln e \\
\Rightarrow 0.693 = 0.3t \\
\Rightarrow t = 2.31 \,hours \;
\]
Thus, our second answer is: The doubling time of the population is \[2.31 \, hours\] .
Note: Here, in this question, we have used two rules of the logarithms. First, the exponent rule: $ \ln {a^b} = b\ln a $ . And the second rule which states that the logarithm of the number with the same base is 1 that is $ \ln e = {\log _e}e = 1 $ .
Formula used:
Systems that experiences exponential growth increase according to the mathematical model:
\[P = {P_0}{e^{kt}}\]
Where, $ {P_0} $ represents the initial state of the system and $ k $ is a constant, called the growth constant.
Complete step by step solution:
We are given that the bacteria colony starts at 100 and grows by \[30\% \] per hour.
Therefore, $ {P_0} = 100 $ and $ k = 0.3 $ .
By using the formula \[P = {P_0}{e^{kt}}\] , we can say that \[P = 100{e^{0.3t}}\] .
Thus, the answer for our first question is: the formula for the number of bacteria, P, after t hours is: \[P = 100{e^{0.3t}}\] .
We also know that the continuous growth rate is nothing but the growth constant.
Therefore, the answer to the third question is: the continuous growth rate for the colony is $ 0.3 $ .
Now, for the second question, we have to find the doubling time.
Therefore, \[P = 2{P_0} = 200\] . Putting this value into the main equation, we will get
\[
200 = 100{e^{0.3t}} \\
\Rightarrow 2 = {e^{0.3t}} \;
\]
Now, we will take logarithms on both the sides.
\[
\Rightarrow \ln 2 = \ln {e^{0.3t}} \\
\Rightarrow \ln 2 = 0.3t\ln e \\
\Rightarrow 0.693 = 0.3t \\
\Rightarrow t = 2.31 \,hours \;
\]
Thus, our second answer is: The doubling time of the population is \[2.31 \, hours\] .
Note: Here, in this question, we have used two rules of the logarithms. First, the exponent rule: $ \ln {a^b} = b\ln a $ . And the second rule which states that the logarithm of the number with the same base is 1 that is $ \ln e = {\log _e}e = 1 $ .
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

