Expert: Josh - 3/27/2010

Question
how to calculate rate of interest using log tables in compound interest

Answer
Hi Umesh,

One way of using logarithm calculation, is to find the period (i.e., how long it takes) to reach a target value, from an initial value given the growth rate.

Suppose we let
| F=future value (money in account after n months)
| P=present value (money in account initially)
| r=interest rate (let's assume it is calculated monthly)
| n=number of months it takes to reach or exceed the target value.

Problem:
Our goal is to reach F=2500, starting from P=1000. Let's say the interest rate is r=0.08 (8 percent per month). How many months (n) does it take to reach our target?

Formulation:
Begin with the compound interest equation, F=P(1+r)^n.
Make "n" the subject of the equation, n=log_{1+r}(F/P) ....[#1]
Using the laws of logarithm, viz.,
| (i)   log(a/b)=log(a)-log(b), and the change of base rule
| (ii)  log_p(y)=log_x(y)/log_x(p)], where x is an arbitrary base (commonly, x=2,10 or e)
[#1] becomes:
n=log_{1+r}(F)-log_{1+r}(P)  using rule (i)
=[log(F)-log(P)]/log(1+r)   where log is logarithm to any base value.

Implication:
Using any log table (e.g., log_10, log_2 etc.), we simply look up 3 values
log(F), log(P) and log(1+r) to work out the value of "n" (how many months it takes).

In this example, F=2500, P=1000, r=0.08.
If we use log_10 for the logarithm, using a look up table (or calculator), we get the following values.
| log(2500) ~= 3.397940
| log(1000) = 3
| log(1.08) ~= 0.033423
Hence, n=(3.397940-3)/0.033423=11.90... (about 12 months)