maths (interest)

Let r be the interest rate.


Initial deposit = $x

Amount = $y

Year = n years

So, x(1 + r)^n = y

(1 + r)^n = y/x

nlog(1 + r) = log(y/x)



So, after 3n years:

Let the amount be a

i.e. x(1 + r)^3n = a

(1 + r)^3n = a/x

3[nlog(1 + r)] = log(a/x)

3log(y/x) = log(a/x)

(y/x)³= a/x

y³/x³ = a/x

a = y³/x²

So, the amount received after 3n years is $ y³/x²