Salmonella bacteria grows rapidly in a nice warm place. If just a few hundred were left on a cutting board when a chicken was cut up, and they get into the potato salad, the population begins compounding. Suppose the number present in the potato salad after x hours is given by: f(x)=500(2^(3x))

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 often do the bacteria double?
d.)How quickly will the number of bacteria increase to 32,000?

(a) 500*2^3 = 500*8 = 4000

(b) 500*2^0 = 500*1 = 500
(c) when 3x=1 or every 1/3 hour
(d)
500*2^3x= 32000
2^3x = 32000/500 = 64
2^3x = 2^6
3x = 6
x = 2

a) To find the number of bacteria present one hour later, substitute x = 1 into the function f(x):

f(1) = 500(2^(3(1))) = 500(2^3) = 500(8) = 4000

Therefore, there are 4000 bacteria present in the potato salad one hour later.

b) To find the initial number of bacteria, substitute x = 0 into the function f(x):

f(0) = 500(2^(3(0))) = 500(2^0) = 500(1) = 500

Therefore, initially there were 500 bacteria present.

c) The bacteria double in number when the exponent of 2 in the function f(x) increases by 1. In other words, with each increase of x by 1, the number of bacteria doubles.

d) To find how quickly the number of bacteria increases to 32,000, we need to solve the equation 32000 = 500(2^(3x)) for x:

32000 = 500(2^(3x))
64 = 2^(3x)
(2^6) = (2^3x)
6 = 3x
x = 2

Therefore, it will take 2 hours for the number of bacteria to increase to 32,000.

To answer these questions, we'll use the given formula for the population of bacteria in the potato salad after x hours: f(x) = 500(2^(3x))

a.) To find the number of bacteria present 1 hour later (x = 1), we substitute x = 1 into the formula:
f(1) = 500(2^(3*1))
f(1) = 500(2^3)
f(1) = 500(8)
f(1) = 4,000

Therefore, 4,000 bacteria would be present in the potato salad 1 hour later.

b.) To determine the initial number of bacteria present, we need to find f(0):
f(0) = 500(2^(3*0))
f(0) = 500(2^0)
f(0) = 500(1)
f(0) = 500

Hence, the initial number of bacteria present is 500.

c.) We find out how often the bacteria double by determining the value of x when f(x) is equal to twice the initial number of bacteria (2 * 500 = 1000):
2 * 500 = 500(2^(3x))
1000 = 2^(3x)

To simplify this equation, we convert 1000 into powers of 2:
1000 = 2^3 * 5^3 = (2 * 2 * 2) * (5 * 5 * 5)
1000 = 2^3 * 5^3 = 2 * 2^2 * 5 * 5 * 5
1000 = 2^3 * 5^3 = 2^(3+2) * 5^3 = 2^5 * 5^3

Comparing this to our equation, we can see that 3x = 5, or x = 5/3.

Therefore, the bacteria double every 5/3 hours.

d.) To determine how quickly the number of bacteria will increase to 32,000, we need to find the value of x when f(x) = 32,000:
32,000 = 500(2^(3x))

Dividing both sides of the equation by 500, we get:
64 = 2^(3x)

To simplify this equation, we express 64 as a power of 2:
64 = 2^6

Comparing this to our equation, we can see that 3x = 6, or x = 6/3.

Therefore, the number of bacteria will increase to 32,000 in 2 hours.