Question #2932b

1 Answer
Somebody N.
Oct 16, 2017

a). = #ln(8)#

b). #y= 55*8^t#

c). 922746880

d). 0.115 (3.d.p)

e). 2.84 (2.d.p.)

Explanation:

We are looking for an exponential equation of the type:

#y= ae^(kt)#

Where:

#a# is the initial amount.(55)

#k# is the growth rate.

#t# is the time. ( In this case hrs )

We know the cells divide every 20 minutes, or 3 times an hr, so it is reasonable to assume that after one hour we have:

#2(2(2xx55))=2^3*55= 440# ( This is the value of #y# after 1 hr )

To calculate growth rate, we plug in the information we now know.

#440= 55e^(k(1))#

(t = 1 hr )

We now solve this for #k#:

# 55e^(k(1))=440#

Dividing by #55# and taking natural logs:

#kln(e) = ln(440/55)= kln(e)=ln(8)#

#ln(e)=1#

So:

#k=ln(8)#

Note: #e^(ln(a))=a#

Expression for number after t hours.

#y= 55*8^t#

Number of cells after 8 hours:

#t = 8#

#y= 55*8^(8)=> y = 922746880#

Rate of growth after 8 hour (billions per hr):

#(922746880/8)* (1/10^9)= 0.1153433600 =0.115# (3.d.p)

Population reaching 20, 000:

#20000= 55*8^t#

#8^t= 20000/55= 4000/11#

#t = (ln(4000/11))/(ln(8)) = 2.835450...= 2.84# (2.d.p.)

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