# Calculating the period of an investment

For calculations using the simple interest formula, we solve for n, the time period of an investment or loan, by simply rearranging the formula to make n the subject. For compound interest calculations, where n is an exponent in the formula, we need to use our knowledge of logarithms to determine the value of n. $A=P(1+i)^n$

A = accumulated amount

P = principal amount

i = interest rate written as a decimal

n = time period

Solving for n: $A=P(1+i)^n$ $\frac{A}{P}=(1+i)^n$

Use deﬁnition : $n=\log_{(1+i)}(\frac{A}{P})$

Change of base $n=\frac{(\log \frac{A}{P})}{\log (1+i)}$

Example 1 :

Thembile invests R 3500 into a savings account which pays 7;5% per annum compounded yearly. After an unknown period of time his account is worth R 4044;69. For how long did Thembile invest his money?

Solution:

Step 1: Write down the compound interest formula and the known values $A=P(1+i)^n$

A = 4044;69

P = 3500

i = 0;075

Step 2: Substitute the values and solve for n $A=P(1+i)^n$ $4044,69=3500(1+0,075)^n$ $\frac{4044,69}{3500}=1,075^n$ $n=\log_{1,075}(\frac{4044,69}{3500})$ $n=\frac{\log \frac{4044,69}{3500}}{\log 1,075}$ $n=2,00...$

Step 3: Write final answer

The R 3500 was invested for 2 years.