Question

Isaac purchased a house for $179,300,00. Every year, Isaac makes improvements so that the value of the house goes up by 4%.

Which of the following equations can be used to determine the number of years after purchase, that the value of Isaac's house will
be equal to 5197.230.002

Answer 1

The number of years in which house value goes up is 145 years .

Explanation:

Given as :

The initial purchased value of the house = p = $179,300,00

The value of house goes up every years at the rate = r = 4%

Let The number of years in which house value goes up = t years

The value of the house after t years = $A = $5197,230,002

Now, According to question

The value of the house after t years = The initial purchased value of the house ×
`(1+(\textrm rate)/(100))^(\textrm time)`

I.e A = $p ×
`(1+(\textrm r)/(100))^(\textrm t)`

Or,$5197,230,002 = $17930000 ×
`(1+(\textrm 4)/(100))^(\textrm t)`

Or,
`(5197,230,002)/(179,300,00)` =
`(1+(\textrm 4)/(100))^(\textrm t)`

Or, 289.86 =
`(1.04)^(t)`

Now, taking Log both side

So,
`Log_(10)`289.86 =
`Log_(10)`
`(1.04)^(t)`

or, 2.462 = t ×
`Log_(10)1.04</p><p>or, 2.462 = t × 0.01703</p><p>∴ t = [tex](2.462)/(0.01703)`

I.e t = 144.56 ≈ 145

So, Number of years = t = 145 years

Hence The number of years in which house value goes up is 145 years . Answer