Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 7.2
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Exercise 136 Page 353

Practice makes perfect
a

We have been given an equation that describes the spread of the virus. Examining the equation, we see that it is an exponential function of the form y=ab^x, where a is the initial value and b is the multiplier. We can identify these variables directly.

y&= 5 000 000( 1/2)^t a&= 5 000 000 b&= 1/2 As we can see, the initial value is 5 000 000. This is the initial number of bytes of information on Deniz's hard drive.
b

By setting y equal to 1000, we can determine the number of minutes it will take, t, for the number of bytes to equal this amount.

y=5 000 000(1/2)^t
1000=5 000 000(1/2)^t
Solve for t
1000/5 000 000=(1/2)^t
1/5000=(1/2)^t
(1/2)^t=1/5000

log(LHS)=log(RHS)

log(1/2)^t=log_(10) 1/5000

log(a^m)= m*log(a)

tlog1/2=log 1/5000
t=log 15000/log 12
t=12.28771...
t≈ 12.29

After about 12.29 minutes there are 1000 bytes of information left.

c

The hard drive will be completely erased when there are 0 files left. However, we are dealing with an exponential function, which means the model will only approach 0. This means we cannot solve y=0. To approximate the number of minutes it will take, we can instead solve the inequality y<1.

y<1
5 000 000(1/2)^t<1
Solve for t
(1/2)^t<1/5 000 000

log(LHS)

log(1/2)^t

log(a^m)= m*log(a)

t* log1/2
t>log 15 000 000/log 12
t>22.25349...

After about 22.25 minutes the number of bytes is less than 1. Therefore, this is the best approximation for when the number of bytes is equal to 0.