A virus sample multiplies itself by becoming \(16\) times itself every hour. But due to anti-agent present in the system

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A virus sample multiplies itself by becoming \(16\) times itself every hour. But due to anti-agent present in the system, only half of them survive every hour. If we start with \(x\) number of virus and after the \(8\) hours there are \(2^{40}\) viruses, what is the value of \(x?\)

A. \(2^{16}\)
B. \(2^{18}\)
C. \(2^{21}\)
D. \(2^{23}\)
E. \(2^{25}\)

Answer: A

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Every hour virus increases by 16 times itself due to the antivirus/anti agent present, only half of them survive


Given that initial value = x
per hour, x increases by x * 16 * 1/2 = x * 8
After shows, there are 2^40 viruses because the virus multiplies itself, the growth is exponential so for 8 hours
$$x\cdot\left(8\right)^8=2^{40}$$
$$x\cdot\left(2^3\right)^8=2^{40}$$
$$x\cdot2^{24}=2^{40}$$
$$x=\frac{2^{40}}{2^{24}}$$
$$x=2^{40-24}$$
$$x=2^{16}$$


$$Answer\ =\ A$$

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VJesus12 wrote:
Tue Sep 15, 2020 5:33 am
A virus sample multiplies itself by becoming \(16\) times itself every hour. But due to anti-agent present in the system, only half of them survive every hour. If we start with \(x\) number of virus and after the \(8\) hours there are \(2^{40}\) viruses, what is the value of \(x?\)

A. \(2^{16}\)
B. \(2^{18}\)
C. \(2^{21}\)
D. \(2^{23}\)
E. \(2^{25}\)

Answer: A

Solution:

If the anti-agent kills half of virus every hour, the virus only multiplies itself 8 times every hour. So if at the start, there are x number of the virus and after 8 hours, there are 2^40 viruses, we can create the equation:

x * 8^8 = 2^40

x * (2^3)^8 = 2^40

x * 2^24 = 2^40

x = 2^16

Answer: A

Scott Woodbury-Stewart
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