3.4  Indirect Use: Example 4

Problem:   Joan places $500 dollars in an investment. Four years later, the investment was worth $645.34. What interest rate, compounded yearly, would give her this amount?

The arithmetic solution given next will be followed by an algebraic solution. The algebraic solution again captures a pattern present in the arithmetic solution.

ARITHMETIC SOLUTION. (Suggestion: look at the shorthand solution first). Here we are given the final amount A = $645.34, the initial amount (principal) P = $500 and the number n = 4 of periods (here years) that the money is earning interest. The compound interest formula says

A = P(1+n)n

With the values given, we may write

$645.34 = $500(1+i)4

From the previous equation, we get two more equalities:

$645.34 $500 = $500(1+i)4 $500 = (1+i)4

The first equality is justified since we can replace $645.34 by its equal $ 500(1+i)⁴ . The second follows since the number (1+i)ⁿ is both multiplied and divided by the quantity $500.

This yields

⁸An equation which always holds is called an identity.

 From both together, we get

Next we use a calculator to see

i = [1.29068][¼]-1 = 1.065870997 -1 = .065780997

approximately. But this is a number. To express the answer as a percentage we note

0.065870997 = (.065870997 ×100)× 1 100 = 6.5870997×0.01 = 6.5870997×1.0%

since 1.0% = .01 = [1/100]. The answer is that the previously unknown and forgotten interest rate i = 6.5870997%.

ALGEBRAIC SHORTHAND SOLUTION. The shorthand solution is as follows. The first part of this solution does not care what values have been acquired by the symbols A, n and P. We are about to solve many problems at once. This shows the power of shorthand notation in manipulating equations.

We will start with the compound interest formula

A = P(1+i)n

Again, if you do not like dealing with letters, suppose A, n and P are shorthand for the numbers given in the arithmetic solution (or any other numbers you want). On your second reading, imagine or give other values for the three quantities in question. As a first step to isolate the interest rate i, we note

A P = P(1+i)n P

upon replacing the quantity A by its equal P(1+i)ⁿ. Now multiplying and dividing by a nonzero quantity P ¹ 0 is the same as multiplying by 1. Therefore the right hand side is just (1+i)ⁿ. This observation suggests the intermediate result

A P = (1+i)n

This gives us two expressions whose values when computed should be equal. Now we take the [1/(n)]-th power (n-th root) of the left-handside [(A)/(P)], and then replace it by its equal. This yields⁹

é ê ë A P ù ú û 1/n   = [(1+i)n]1/n = 1+i

⁹When a is a positive number, b = a^[1/(n)] stands for the number b > 0 which satisfies bⁿ = a.

The examples above show the role of algebraic shorthand notation in describing and in changing the way calculations are done. The compound interest formula has been used and manipulated but the origin of this formula has not been described. Ask a mathematics instructor for an explanation.

More Chapter Sections: [ Up ] [ 14 The Formula ] [ 14. Direct Use - First Example ] [ 14. Direct Use, Second Example ] [ 14 Indirect Use - First Example I ] [ 14 Indirect Use - Second Example ] [ 14 Going Further ]