A pathologist, trying to identify a bacterial infection, places a sample in a growth chamber and measures

1000 bacteria at the end of day 5 of the test. From daily measurements, the growth rate of the number of
bacteria is determined to be 5% per day.

Note that:

A linear relationships is of the form y = A + kx

A logarithmic/exponential relationship is of the form y = Aekx

A power law relationship is of the form y = Axk

What type of graph will give a straight line plot of the number of bacteria vs. time?

Determine A and k for that form.

How many bacteria were in the sample when the test started at the beginning of day 1?

To determine the type of graph that will give a straight line plot of the number of bacteria vs. time, we need to consider the form of the growth equation.

The given growth rate of the number of bacteria is determined to be 5% per day. This indicates an exponential growth relationship, where the number of bacteria increases by a constant factor (5% in this case) each day.

The exponential growth relationship is of the form y = A * e^(kx), where y represents the number of bacteria, x represents time, A is the initial quantity of bacteria, and k is the growth rate.

In this case, the number of bacteria (y) is given, and we need to determine A and k.

To find A, we can use the formula:

A = y / e^(kx)

For the given data point at the end of day 5, with 1000 bacteria, we can plug in the values:

A = 1000 / e^(5k)

To find k, we need to use the growth rate of 5% per day. The growth rate (k) is determined by finding the natural logarithm of (1 + growth rate).

k = ln(1 + growth rate) = ln(1 + 0.05) = ln(1.05)

Now that we have the value of k, we can substitute it into the equation for A:

A = 1000 / e^(5ln(1.05))

Simplifying this expression will give us the value of A.

To determine the number of bacteria in the sample at the beginning of day 1, we need to substitute the values of A and k into the exponential growth equation with x = 0 (since we are interested in the initial quantity).

y = A * e^(k * 0) = A * e^0 = A

Therefore, the number of bacteria in the sample at the beginning of day 1 is equal to A.

To summarize:

- The graph that will give a straight line plot of the number of bacteria vs. time is a logarithmic/exponential graph with y = A * e^(kx).
- The values of A and k can be determined by substituting the given data into the formulas mentioned above.
- The number of bacteria in the sample at the beginning of day 1 is equal to the value of A.