P'(t) = 94000*0.03 e^0.03t = 2820 e^.03t
so, in 2000, the value will be rising at
2820 e^.3 = $3807/yr
to find the doubling interval, solve for t:
2 = e^.03t
t = ln2/.03 = 23.10 years
The price in dollars of a house during a period of mild inflation is described by the formula P(t)=94000 e0.03 t, where t is the number of years after 1990. Answer the following questions:
B. In the year 2000 the value will be increasing at a rate of dollars per year. (Round your answer to the nearest dollar.)
C. How long will it take for a house to double in value? Answer: years. (Round your answer to two decimal places.)
2 answers
for C.
In 1990, the house is worth 94,000 to double it has to be worth 188,000
Putin 188,00 for P(t) then divide both sides by 94,000. That will give you
2= e^.03t
take the ln of both sides then solve for t.
Are you in calculus? If so.. for B. I would take the derivative that will give you rate of change and then put in the t which will be 10 years.
In 1990, the house is worth 94,000 to double it has to be worth 188,000
Putin 188,00 for P(t) then divide both sides by 94,000. That will give you
2= e^.03t
take the ln of both sides then solve for t.
Are you in calculus? If so.. for B. I would take the derivative that will give you rate of change and then put in the t which will be 10 years.