Question

A farm increases in value from $800,000 to $1,100,000 over a period of 6 years. use the formula r=(fp)1/n−1r=(fp)1/n−1 to find the annual inflation rate rr to the nearest tenth of a percent, where nn is the number of years during which the value increases from pp to ff.

Answer 1

Future value is given by;
f = p(1+r)^n
f = future value = $1,100,00
p = present value = $800,000
r = inflation rate
n = number of years in which the vale increases from p to f = 6

Therefore,
1,100,000 = 800,000 (1+r)^6
1100000/800000 = (1+r)^6
1.375 = (1+r)^6
log (1.375) = 6 log (1+r)
[log(1.375)]/6 = log (1+r)
0.0231 = log (1+r)
e^0.0231 = 1+r
1.0545 = 1+r
r= 1.0545 - 1 = 0.0545 = 5.45%

The rate of inflation is 5.45%.