Question

Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year. If you had purchased a house in Midvale, UT for $ 172000 in 2004 in what year would the home be worth $ 249000 ?

Answer 1

The home would be worth $249,000 in the year 2012.

Step-by-step explanation:

To calculate the year when the home would be worth $249,000, we need to use the formula for exponential growth:

Final Value = Initial Value * (1 + Growth Rate)Time

Given that the initial value is $172,000, the growth rate is 4.7% (or 0.047), and we need to find the time when the home is worth $249,000, we can plug in these values into the formula and solve for Time:

$249,000 = $172,000 * (1 + 0.047)Time

Taking the logarithm of both sides, we get:

log(1 + 0.047)($249,000 / $172,000) = Time

Using a scientific calculator or an online logarithm calculator, we can find the value of Time, which should be approximately 8.32 years.

Therefore, the home would be worth $249,000 in the year 2004 + 8.32 = 2012.

Answer 2

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:

`P(t) = P(0)(1+r)^(t)`

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
`r = 0.047`

$172000 in 2004

This means that
`P(0) = 172000`

What year would the home be worth $ 249000 ?

t years after 2004.

t is found when P(t) = 249000. So

`P(t) = P(0)(1+r)^(t)`

`249000 = 172000(1.047)^(t)`

`(1.047)^(t) = (249000)/(172000)`

`\log{(1.047)^(t)} = \log{(249000)/(172000)}`

`t\log(1.047) = \log{(249000)/(172000)}`

`t = \frac{\log{(249000)/(172000)}}{\log(1.047)}`

`t = 8.05`

2004 + 8.05 = 2012

The home would be worth $249000 during the year of 2012.