Question

If monthly payments p are deposited in a savings account paying an annual interest rate r, then the amount A in the account after n years is given by

A= p(1+r/12)[(1+r/12)^12n - 1]/ (r/12)

Graph A for the values of p and r.
p = 100, r = 0.06

Estimate n for A = $200,000. (Round your answer to one decimal place.)

Answer 1

r/12 = .06/12 = .005
1+r/12 = 1.005

A = 100 (1.005) [ 1.005^12n -1]/.005

200,000 = 20,000 (1.005)[ 1.005^12n -1]

2000 = [ 1.005^12n -1]

2001 = 1.005^12n

log 2001 = 12 n log 1.005

12 n = 1524.08

n = 127 years
check my arithmetic !