A house was valued at $120,000 in the year 1992. The value appreciated to $160,000 by the year 2007

Question

Use the compund interest formula S=P(1+r)t to answer the following questions.

A) What was the annual growth rate between 1992 and 2007?

r =____ Round the growth rate to 4 decimal places.

B) What is the correct answer to part A written in percentage form?

r = _____%.

C) Assume that the house value continues to grow by the same percentage. What will the value equal in the year 2010 ?

value = $ ______ Round to the nearest thousand dollars.

Answer 1

S=P(1+r)^t

A) Between 1992 and 2007 is 15 years, so t = 15

160000 = 120000(1+r)¹⁵ Divide both sides by 120000

160000/120000 = 120000/120000(1+r)¹⁵ Simplify

4/3 = (1+r)¹⁵ Raise both sides to the 1/15 to eliminate the exponent

(4/3)^1/15 = ((1+r)¹⁵)^1/15 Simplify

1.0194 = 1+r Subtract 1 from both sides

0.0194 = r

B) To convert to a percentage multiply by 100

0.0194(100) = 1.94%

C) t = 18 years for 2010

S=120000(1+.0184)¹⁸

S=$169,584.88