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.

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%.