Answer:
The home would be worth $249000 during the year of 2012.
Step-by-step explanation:
The price of the home in t years after 2004 can be modeled by the following equation:
[tex]P(t) = P(0)(1+r)^{t}[/tex]
In which P(0) is the price of the house in 2004 and r is the growth rate.
Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year.
This means that [tex]r = 0.047[/tex]
$172000 in 2004
This means that [tex]P(0) = 172000[/tex]
What year would the home be worth $ 249000 ?
t years after 2004.
t is found when P(t) = 249000. So
[tex]P(t) = P(0)(1+r)^{t}[/tex]
[tex]249000 = 172000(1.047)^{t}[/tex]
[tex](1.047)^{t} = \frac{249000}{172000}[/tex]
[tex]\log{(1.047)^{t}} = \log{\frac{249000}{172000}}[/tex]
[tex]t\log(1.047) = \log{\frac{249000}{172000}}[/tex]
[tex]t = \frac{\log{\frac{249000}{172000}}}{\log(1.047)}[/tex]
[tex]t = 8.05[/tex]
2004 + 8.05 = 2012
The home would be worth $249000 during the year of 2012.
