Let's solve the problem:

Given:

• The bacteria colony doubles every 6 hours.
• We need to find out how many hours it will take for the bacteria to be 5 times the original amount.

This is a typical exponential growth problem. We can use the following formula for exponential growth:

$N(t) = N_0 \cdot 2^{\frac{t}{T}}$

Where:

• $N(t)$ is the number of bacteria at time $t$,
• $N_0$ is the initial number of bacteria,
• $T$ is the doubling time (6 hours in this case),
• $t$ is the time elapsed in hours.

We are asked to find $t$ when the bacteria is 5 times the original amount, i.e., $N(t) = 5N_0$. Substituting into the equation:

$5N_0 = N_0 \cdot 2^{\frac{t}{6}}$

Dividing both sides by $N_0$ (since $N_0 \neq 0$):

$5 = 2^{\frac{t}{6}}$

Now, take the logarithm (base 2) of both sides:

$\log_2(5) = \frac{t}{6}$

Solving for $t$:

$t = 6 \cdot \log_2(5)$

We can convert the base 2 logarithm into base 10 (or natural logarithm):

$t = 6 \cdot \frac{\log(5)}{\log(2)}$

Let me calculate this for you.The correct time it will take for the bacteria to become 5 times the original amount is approximately 13.932 hours, rounded to three decimal places.

It looks like the answer you provided, 13.931, was very close but slightly off due to rounding.

Would you like further details or have any questions about the calculation?

Here are some related questions:

1. What is the general formula for exponential growth and decay?
2. How can we calculate the time for different multiplication factors (e.g., 3 times, 10 times)?
3. How does changing the doubling time affect the growth?
4. What other natural processes follow exponential growth patterns?
5. How would the calculation change if the doubling time were 8 hours instead?

Tip: Always check for rounding errors when precision is required to a specific decimal point!