Question

The tuition in the school year 2012–2013 at a certain university was $13,000. For the school year 2017–2018, the tuition was $14,690. Find an exponential growth function for tuition T (in dollars) at this university t years after the 2012–2013 school year. (Round your values to four decimal places.)T = _______Assuming it increases at the same annual rate, use the function to predict the tuition (in dollars) in the 2021–2022 school year. (Round your answer to the nearest integer.)$________

Answer 1

$\begin{gathered} a)\text{ }T=13000(1.0247)^t \\ b)\text{ \ 16193} \end{gathered}$

Explanation:

Tuiton from 2012 - 2013 = $13000

Tuition from 2017-2018 = $14690

Time difference = t= 2017 - 2012 = 5 years interval

We apply exponential function formula:

$\begin{gathered} y=ab^x \\ a\text{ = 13000 = inital tuition} \\ b\text{ = ? , }x\text{ = number of years }=\text{ t} \\ t\text{= 5} \\ y\text{ = T} \\ y\text{ = 14690} \\ 14690=13000b^5 \end{gathered}$$\begin{gathered} \frac{14690}{13000}=b^5 \\ 1.13=b^5 \\ b\text{ = }\sqrt[5]{1.13} \\ b\text{ = 1.02}47 \\ \\ \text{Hence, the exponetial growth function becomes:} \\ T=13000(1.0247)^t\text{ (4 decimal place)} \end{gathered}$$\begin{gathered} \text{ Tuition fe}e\text{ for year 2021-2022:} \\ We\text{ n}eed\text{ to find the time difference of school year 2012-2013 from 2021-2022} \\ 2021\text{ - 2012 = 9} \\ 2022\text{ - 2013 = 9} \\ t\text{ = 9 years} \\ \text{From the function gotten above: same rate} \\ T=13000(1.0247)^t \\ T=13000(1.0247)^{9^{}} \\ T\text{ = }16192.5027 \\ \\ To\text{ the nearest integer, T = \ }16193 \end{gathered}$