# Salmonella bacteria, found on almost all chicken and eggs,grow rapidly in a nice warm place. If...

## Question:

Salmonella bacteria, found on almost all chicken and eggs,grow rapidly in a nice warm place. If just a few hundred bacteria are left on the cutting board when a chicken is cut up,and they get into a potato salad, the population begins compounding. Suppose the number present in the potato salad after {eq}x {/eq} is given by {eq}f(x) = 500 \cdot 2^{3x} {/eq}

a. If the potato salad is left out on the table, how many bacteria are present 1 hour later?

b. How many were present initially?

c. How after do the bacteria double?

d. How quickly will the number of bacteria increase to 32,000?

## Exponential Functions

An exponential function is written in the form {eq}y = C \cdot a^{kx} {/eq}, where C is the initial value of the quantity and k is the growth or decay rate, depending on its sign and the value of a. If a is 2 and k is positive, we can use it to determine the doubling time of the quantity. This is is because the doubling time is then equal to {eq}\frac{1}{k} {/eq}.

## Answer and Explanation:

a. Let's evaluate this function at 1 to find how much bacteria is in this population after 1 hour.

{eq}\begin{align*} f(1) &= 500 \cdot 2^{3(1)} \\ &= 500(8)\\ &= 4000 \end{align*} {/eq}

b. The initial value of an exponential function is defined as the coefficient in front of the exponential base. Therefore, this population begins as 500 bacteria. If we weren't convinced with this definition, we could evaluate this function at zero to arrive at this same conclusion.

c. The doubling time of a population can be found as the constant d if our function is in the form {eq}y = C\cdot 2^{\frac{t}{d}} {/eq}. If we rearrange our function, we would need to do something about the constant 3 in the exponent. This would mean that we need to rewrite our function where d is the reciprocal of 3: {eq}f(x) = 500 \cdot 2^{\frac{t}{1/3}} {/eq}. This means that the doubling time of this population is one-third hour.

d. We can set our function equal to a specific population and solve for the amount of time it would take for the population to reach that size using logarithms.

{eq}32000 = 500 \cdot 2^{3x}\\ 64 = 2^{3x}\\ \log_2 64 = \log_2 2^{3x}\\ 3x = 6\\ x = 2 {/eq}

It therefore takes only two hours for the population to reach this size.

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10