Suppose that an initial population of 7000 bacteria grows exponentially at a rat

Question

Suppose that an initial population of 7000 bacteria grows exponentially at a rate of 4% per hour and that y = y(t) is the number of bacteria present t hours later.

(c) How long does it take for the initial population of bacteria to double?

Round your answer to three decimal places.

t ~ h

(d) How long does it take for the population of bacteria to reach 47,000?

Round your answer to two decimal places.

Suppose that an initial population of 7000 bacteria grows exponentially at a rate of 4% per hour and that y = y(t) is the number of bacteria present t hours later.

Explanation / Answer

the exponential function is given as :

A=yb^t

where A is the number of bacteria after time t

y is the initial number of bacteria

and b= 1+r =1+.04=1.04 is the growth factor

=> at t=1

A=7000(1.04)^1 = 7280

now use

y= yo e^kt

7280 = 7000e^(k(1))
1.04 = e^k
ln1.04 = lne^k
k = .0392
y = 7000e^0.0392t
y' = (0.0392)(7000)e^0.0392t

c> y = 7000e^0.0392t

y = 14000

=> 14000=7000e^(.0392t)

=> t= 17.682 hours

d>47000 = 7000e^(.0392t)

=> t = 48.58 hours

y' = .0392yo e^(kt) = 0.0392y

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