Question 278791
formula is:


f = p * (1+i)^n


f = future value
p = present amount
i = annual interest rate compounded annually
n = number of time periods in years


you have:


f = 5000
p = 2000
i = i
n = 14


your formula becomes:


5000 = 2000 * (1+i)^14


divide both sides of this formula by 2000 to get:


5000/2000 = (1+i)^14


Take the log of both sides of this equation to get:


log(5000/2000) = log(1+i)^14)


since log (x^a) = a*log(x), your formula becomes:


log(5000/2000) = 14 * log(1+i)


divide both sides of this equation by 14 to get:


log(1+i) = log(5000/2000)/14


Use your calculator to find tghe log of (5000/2000) and you get:


log(1+i) = .397940009/14


Simplify to get:


log(1+i) = .028424286


Use your calculator to find the number whose log is .028424286 and you get:


1+i = 1.067638647


Subtract 1 from both sides of this equation to get:


i = .067638647 which is equivalent to 6.7638647%


Your annual interest rate is .067638647 which is equivalent to 6.7638647% per year when expressed as a percent.


Plug this rate (not the percent) back into your original equation, which is:


5000 = 2000 * (1+i)^14


replace i with .067638647 to get:


5000 = 2000 * (1.067638647)^14


Simplify to get:


5000 = 2000 * 2.5 = 5000 confirming that the interest rate of .067638647 is good.