On a day, a video was posted online, 5 people watched the video. The next day the number of viewers had doubled. This tend continues each day. Write an equation modeling the views, v, as a function of time.

On which day will 640 people see the video? Explain if you can.

Step-by-step explanation:

The number of viewers is a geometric sequence with 5 viewers on day 1 and a common ratio of 2. The general term will be ...

  v(t) = 5·2^(t-1) . . . . . . viewers on day t

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We want to find t for v(t) = 640.

  640 = 5·2^(t-1) . . . . . use v(t)=640

  128 = 2^(t-1) . . . . . . . divide by 5

  2^7 = 2^(t -1) . . . . . . . recognize 128 = 2^7

  7 = t -1 . . . . . . . . . . . . equate exponents of 2

  8 = t . . . . . . . add 1

There will be 640 viewers on day 8.

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Comment on the equation and solution

The problem statement doesn't tell you how time is counted. If you treat the problem as one requiring an exponential function, you can also write the equation as ...

  v(t) = 5·2^t

where t is the number of days after the first day. Then the solution will be ...

  640 = 5·2^7   ⇒   t = 7 . . . . days after the first day

7 days after the first day is the same as Day 8, if the first day is Day 1.

step-by-step explanation:

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).