Question

I don't understand question 4.

Answer 1

Your answer for A is correct.

For B, I would keep the base of the logarithm as simple as possible.

`y = 100 \cdot 3.24^x \implies \frac y{100} = 3.24^x \implies \log_(3.24)\left(\frac y{100}\right) = \log_(3.24)(3.24)^x \\\\ \implies x = \log_(3.24) \left(\frac y{100}\right)`

In part A, the function told us the number of students with cell phones (dependent variable) after some number of years post-2000 (independent variable). Our new function turns this around and tells us the time in years after 2000 (dependent) based on how many student have cell phones (independent).

For C, if we set
`y=7200` students, then we can estimate the time it takes for the all students to acquire cell phones by solving for
`x`.

`x = \log_(3.24)\left((7200)/(100)\right) = \log_(3.24)(72) \approx 3.63794`

If you're using a typical calculator, it probably won't have buttons for logarithms of any base other than 10 or the natural base
`e`. To work around that, use the change-of-base identity

`\log_(3.24)(72) = (\log_(10)(72))/(\log_(10)(3.24)) = (\ln(72))/(\ln(3.24))`

At any rate, it would take about 3.64 years for all students to get a cell phone, according to this model.