Answer:
In 1981 was the building worth double it’s value.
Step-by-step explanation:
Given : A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation : [tex]A=80000(1.12)^t[/tex], where t represents time in years.
To find : In what year was the building worth double it’s value in 1975?
Solution :
The amount is $80,000.
The building worth double it’s value in 1975.
i.e. amount became A=2(80000).
Substitute in the model,
[tex]2(80000)=80000(1.12)^t[/tex]
[tex](1.12)^t=\frac{2(80000)}{80000}[/tex]
[tex](1.12)^t=2[/tex]
Taking log both side,
[tex]t\log (1.12)=\log 2[/tex]
[tex]t=\frac{\log 2}{\log (1.12)}[/tex]
[tex]t=6.11[/tex]
i.e. Approx in 6 years.
So, 1975+6=1981
Therefore, in 1981 was the building worth double it’s value.
