Expert: Josh - 8/22/2008

Question
principle=$400000,compound interest=$53600  ,rate=6.5        then time=?

Answer
deepak,

we make use of the compound interest equation in this problem.
observe that the amount at time t is given by
a(t)=a(t-1)*(1+r), where a(t-1) represents the amount at the previous time instance t-1.

by induction, a(t)=a(0)*(1+r)^t, where a(0) is the initial amount (viz., the principle)

The accumulated interest at some future time t is the difference between a(t) and a(0).

in this question, the parameters are given by
principle: a(0)=400000,
compound interest: a(t)-a(0)=53600
interest rate: r=(6.5/12)/100 [assuming interest quoted per annum and calculated monthly]

thus, we substitute a(t)=a(0)*(1+r)^t in a(t)-a(0)=53600 and solve for t.

a(0)*[(1+r)^t-1] = 53600
(1+r)^t-1 = 53600/a(0)
(1+r)^t = 53600/a(0) + 1 ....take log base (1+r) to find the exponent
t=log_(1+r){53600/400000 + 1}

hint: [change of log rule] log_y(x) may be computed as log_a(x)/log_a(y) for arbitrary base "a"; for convenience, we may pick a to be 10. just to make this clear, log_(1+r)(x) = log10(x)/log10(1+r).

i'll let you finish this yourself.
the answer is roughly 23 months