The rate of increase in the number of
bacteria in a certain bacteria culture is
proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of
bacteria to be 10 times the number of initial
present.

This problem involves exponential growth, where the rate of change in the population of bacteria is proportional to the current population. The general formula for exponential growth is:

\[
N(t) = N_0 e^{kt}
\]

Where:
- \(N(t)\) is the number of bacteria at time \(t\),
- \(N_0\) is the initial number of bacteria,
- \(k\) is the growth constant,
- \(t\) is the time in hours.

### Step 1: Finding the growth constant \(k\)
We are told that the number of bacteria triples in 5 hours. This means at \(t = 5\), \(N(5) = 3N_0\). Substituting this into the exponential growth equation:

\[
3N_0 = N_0 e^{5k}
\]

Dividing both sides by \(N_0\):

\[
3 = e^{5k}
\]

Taking the natural logarithm (ln) of both sides:

\[
\ln(3) = 5k
\]

\[
k = \frac{\ln(3)}{5}
\]

Now, let's calculate \(k\):

\[
k \approx \frac{1.0986}{5} = 0.2197
\]

### Step 2: Finding the number of bacteria after 10 hours
Now that we know \(k\), we can find the number of bacteria after 10 hours. The equation is:

\[
N(10) = N_0 e^{0.2197 \times 10}
\]

Simplifying:

\[
N(10) = N_0 e^{2.197}
\]

Using the value of \(e^{2.197} \approx 9.0\):

\[
N(10) = 9N_0
\]

Thus, after 10 hours, the number of bacteria will be **9 times** the initial population.

### Step 3: Finding the time when the population is 10 times the initial population
We need to find the time \(t\) such that \(N(t) = 10N_0\). Substituting into the growth equation:

\[
10N_0 = N_0 e^{0.2197t}
\]

Dividing by \(N_0\):

\[
10 = e^{0.2197t}
\]

Taking the natural logarithm of both sides:

\[
\ln(10) = 0.2197t
\]

\[
t = \frac{\ln(10)}{0.2197}
\]

Since \(\ln(10) \approx 2.3026\), we can calculate \(t\):

\[
t \approx \frac{2.3026}{0.2197} \approx 10.48 \text{ hours}
\]

Thus, it will take approximately **10.48 hours** for the population to become 10 times the initial number.

### Summary:
1. After 10 hours, the population will be **9 times** the initial population.
2. It will take approximately **10.48 hours** for the population to become 10 times the initial number.

Would you like more details or have any questions?

Here are 5 related questions to explore this topic further:
1. What happens to the population if the bacteria's growth rate doubles?
2. How does the initial number of bacteria affect the time for tripling?
3. What if the bacteria's growth slows down, how would the model change?
4. How would you calculate the population after half a day (12 hours)?
5. Can this model be applied to real-life bacterial cultures with resource limitations?

**Tip:** In problems involving exponential growth, always check if the rate or time units need to be adjusted for the most accurate calculation.

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Explore how exponential growth applies to bacterial populations. Learn how the population triples in 5 hours, calculate the number of bacteria after 10 hours, and find the time for the population to become 10 times the initial count. This problem covers logarithms and growth constants, ideal for students in Grades 10-12.

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