Question

A magazine currently has 8700 subscribers for its online web version. It is adding members at the rate of R(t) = 190e0.03t subscribers per month. If the proportion of members who remain subscribed t months from now is S(t) = e−0.06t, how many online subscribers will the magazine have three years from now? (Round your answer to the nearest integer.)

Answer 1

Number of subscriber the magazine will have after 3 years from now approximately be 8767

Solution:

Given that magazine currently has 8700 subscribers for its online web version

$\begin{array}{l}{\mathrm{R}(\mathrm{t})=190 \mathrm{e}^{0.03 \mathrm{t}} \text { subscribers/month }} \\\\ {\mathrm{S}(\mathrm{t})=\mathrm{e}^{-0.06 \mathrm{t}}}\end{array}$

After 3 years, time(t) = 36 month

Total number of subscribers after 3 years from now :

Substitute "t" = 36

$\begin{array}{l}{\mathrm{R}(36)=190 \mathrm{e}^{0.03 \times(36)}=190 \times(2.944)} \\\\ {\mathrm{R}(36) \approx 560} \\\\ {\mathrm{S}(36)=\mathrm{e}^{-0.06 \times(36)}=0.12}\end{array}$

Subscribers remaining = 0.12 x 560 = 67.2

The magazine currently has 8700 subscribers

Added Subscriber = 8700 + 560 = 9260

Remaining Subscriber = 8700 + 67.2 = 8767.2

Therefore number of subscriber the magazine will have after 3 years from now approximately be 8767