Question

In a culture, the bacteria count is 1,00,000 initially. The number increases by 10% in the first 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

• A. \( \frac{2 \log 11}{\log 2} \)
• B. \( \frac{\log(11/10)}{\log 2} \)
• C. \( \frac{2 \log 2}{\log 11} \)
• D. \( \frac{\log 2}{\log(11/10)} \)

Hint:

When solving problems involving exponential growth, use the formula \( N(t) = N_0 e^{kt} \) and solve for the unknowns step by step, taking logarithms where necessary.

Answer 1

The Correct Option is A

Step 1: Understanding the problem.
We are given that initially the bacteria count is 1,00,000, and it increases by 10% in the first 2 hours. The rate of growth of the bacteria is proportional to the number of bacteria present. This implies that the bacteria population follows an exponential growth model, which can be expressed as:
\[ N(t) = N_0 e^{kt}, \]
where:
- \( N(t) \) is the population at time \( t \),
- \( N_0 \) is the initial population,
- \( k \) is the constant of proportionality,
- \( t \) is the time in hours.

Step 2: Substituting initial values.
From the problem, we know that initially the bacteria count is 1,00,000, so:
\[ N_0 = 100000. \]
Also, after 2 hours, the bacteria count increases by 10%, so:
\[ N(2) = 1.1 \times 100000 = 110000. \]
Using the exponential growth model, we can substitute these values into the equation for \( N(t) \):
\[ 110000 = 100000 e^{2k}. \]

Step 3: Solving for \( k \).
Now, solve for \( k \) by dividing both sides by 100000:
\[ 1.1 = e^{2k}. \]
Take the natural logarithm of both sides: \[ \ln(1.1) = 2k. \]
Thus: \[ k = \frac{\ln(1.1)}{2}. \]

Step 4: Finding the time when the count reaches 2,00,000.
Now, we need to find the time \( t \) when the bacteria count reaches 2,00,000. Using the exponential growth model again, we have:
\[ 200000 = 100000 e^{kt}. \]
Dividing both sides by 100000: \[ 2 = e^{kt}. \]
Taking the natural logarithm of both sides:
\[ \ln(2) = kt. \]
Substitute the value of \( k \) from
Step 3:
\[ \ln(2) = \frac{\ln(1.1)}{2} \times t. \]
Solve for \( t \): \[ t = \frac{2 \ln(2)}{\ln(1.1)}. \]

Step 5: Final expression.
We can now express the answer as:
\[ t = \frac{2 \log 11}{\log 2}. \]
Final Answer:
Thus, the time when the bacteria count reaches 2,00,000 is:
\[ \boxed{\frac{2 \log 11}{\log 2}}. \]