Find how many years it would take for an investment of $4000 to grow to $7800 at an annual interest rate of 7.9% compounded weekly.

Answer 1

A = P (1 + r/n) ^nt

where A = amount (A = $7800)

P = principal (P = $4000)

r = interest rate (r = 7.9% = .079)

n = number of times interest is compounded annually (n = 52 for weekly compounding)

t = time in years

Solve for t using logarithm rules:

A = P (1 + r/n) ^nt

A / P = (1 + r/n) ^nt

log(A/P) = log(1 + r/n) ^nt

log(A/P) = nt * log(1 + r/n)

nt = log(A/P) / log(1 + r/n)

t = log(A/P) / [n* log(1 + r/n)]

Plug in values and solve for t:

t = log(A/P) / [n * log(1 + r/n)] = log($7800 / $4000) / [52 * log(1 + .079/52)] = 8.46 years