Chapter ISG, Problem 7.3.6P

To determine

To Find:

1. Equation to relate the number of visits to the number of days the promotion is running.
2. Number of days to increase the traffic to 12288.
3. Average daily percentage increase.

Answer to Problem 7.3.6P

1. $v=96\times 2^{\frac{d}{5}}$
2. 35
3. 15%

Explanation of Solution

Given:

Number of visits without advertisement is 96. After starting promotions, the number of visits gets double every 5 days.

The traffic increased to 12288.

Formula used:

$A=A_0{b}^{\frac{t}{c}}$

Where $A$ is the value of the function.

$A_0$ is the initial value

$b$ is the growth factor

$t$ is the time taken

$c$ is the time taken for growth $b$

Using the above formula:

$A_0$ = 96

$b$ =2

$t$ is given as $d$

$c$ = 5 days

1. Therefore, the equation is:

$v=96\times 2^{\frac{d}{5}}$

2. Given that, $v=12288$

Using the above equation:

$\begin{array}{l}12288=96\times 2^{\frac{d}{5}}\\ \Rightarrow 128=2^{\frac{d}{5}}\\ \Rightarrow d=35days\end{array}$

3. Percent Average Daily increase:

$\frac{96\times 2^{\frac{d+1}{5}}-96\times 2^{\frac{d}{5}}}{96\times 2^{\frac{d}{5}}}\times 100=2^{\frac{1}{5}}\times 100\approx 15\%$