# The number y of hits a new website receives each month can be modeled by y = 4060e^kt, where t...

## Question:

The number y of hits a new website receives each month can be modeled by y = 4060e{eq}^{kt} {/eq}, where t represents the number of months the website has been operating. In the website's third month, there were 9,000 hits.

Find the value of k. (Round answer to four decimal places.)

## Hits!

A hit is a terminology that is frequently seen on a website. It basically refers to the number of files downloaded from a website. These include photos, videos, documents, and applications.

Given:

{eq}t=3\\[0.2cm] y(3)=9000\\[0.2cm] {/eq}

We rearrange the equation to find the value of k.

$$y=4060e^{kt}\\[0.2cm] \dfrac{y}{4060}=e^{kt}\\[0.2cm] \ln \left| \dfrac{y}{4060} \right| = \ln \mid e^{kt} \mid \\[0.2cm] \ln \left| \dfrac{y}{4060} \right| =kt\\[0.2cm] k=\dfrac{\ln \left| \frac{y}{4060} \right|}{t}\\[0.2cm]$$

We now substitute the values of {eq}y {/eq} and {eq}t {/eq}.

$$k=\dfrac{\ln \left| \frac{9000}{4060} \right|}{3}\\[0.2cm] k\approx \dfrac{0.796}{3}\\[0.2cm] \color{red}{k\approx 0.265}\\[0.2cm]$$

Therefore, the equation becomes

$$y(t)=4060e^{0.265t}\\[0.2cm]$$