
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
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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 $ .
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