A biologist is researching a newly discovered species of bacteria. At time t=0 hours, she puts on hundred bacteria into a Petri dish. Six hours later, she measures 450 bacteria. Assuming exponential growth, what is the growth rate "k" for the bacteria? (Round your percent to two decimals and include units)

450 = 100 e^(kt)

4.5 = e^(6k)

ln(4.5) = 6 k

k = [ln(4.5)] / 6

Well, it sounds like these bacteria are certainly multiplying in a hurry! Let's see if we can calculate that growth rate "k" for you.

To figure out the growth rate "k" using exponential growth, we can use the formula:

Nt = N0 * e^(kt)

Where:
- Nt is the final population size (450 bacteria)
- N0 is the initial population size (100 bacteria)
- e is Euler's number, approximately 2.71 (a real math superstar, let me tell you!)
- k is the growth rate we're trying to find
- t is the time period in hours (6 hours)

To make things a bit easier, let's first rearrange the formula:

k = ln(Nt/N0) / t

Now, plug in all the values:

k = ln(450/100) / 6

k = ln(4.5) / 6

After doing some quick calculations (and sparing you the long logarithmic explanation), it looks like the growth rate "k" for these bacteria is approximately 0.141% per hour.

So, there you have it! The bacteria are multiplying at an approximate rate of 0.141% per hour. Keep an eye out for any tiny bacterial parties happening in that Petri dish!

To find the growth rate "k" for the bacteria, we can use the exponential growth formula:

N = N0 * e^(kt),

where N is the final population size, N0 is the initial population size, e is the base of the natural logarithm (approximately 2.71828), k is the growth rate, and t is the time.

In this case, we know that the initial population size is 100 bacteria (N0), the final population size is 450 bacteria (N), and the time is 6 hours.

Plugging the values into the formula, we get:

450 = 100 * e^(k * 6).

To isolate k, we can divide both sides of the equation by 100:

4.5 = e^(6k).

Now, take the natural logarithm of both sides of the equation to eliminate the exponential:

ln(4.5) = ln(e^(6k)).
ln(4.5) = 6k * ln(e).

Since ln(e) is equal to 1, the equation simplifies to:

ln(4.5) = 6k.

Finally, we can solve for k by dividing both sides of the equation by 6:

k = ln(4.5) / 6.

Using a calculator, we find k ≈ 0.1604.

Rounding to two decimal places, the growth rate "k" for the bacteria is approximately 0.16 per hour.

To determine the growth rate "k" for the bacteria, we can use the exponential growth formula:

N(t) = N0 * e^(kt)

Where:
N(t) = the population size at time t
N0 = initial population size at time t=0
e = Euler's number (approximately 2.71828)
k = growth rate
t = time elapsed

Given that the initial population size at t=0 is 100 bacteria and the population size after 6 hours is 450 bacteria, we can plug these values into the formula to solve for "k."

450 = 100 * e^(6k)

Divide both sides of the equation by 100:

4.5 = e^(6k)

To isolate "k," take the natural logarithm of both sides of the equation:

ln(4.5) = 6k

Divide both sides of the equation by 6:

k = ln(4.5) / 6

Using a calculator, we can compute this value:

k ≈ 0.227 to two decimal places.

Therefore, the growth rate "k" for the bacteria is approximately 0.23 per hour.