MATH 1030: SAVINGS AND INVESTMENT PLANS
TIM CARSTENS
Section 4C: Savings Plans and Investments Consider the following problem: Problem. Mary puts 180 into a bank account each month. The account earns 2.25% interest, compounded monthly. How much money will be in her account after 5 years? This situation may remind us of a related problem: given a loan, what will be the monthly payments? These two problems are not the same, though. For instance, when paying off a loan, at first the amount owed is very large, so that the interest will be worth a lot of money. However, when saving money, the initial amount saved is very small, so that the interest won’t be worth very much at all. So while we’d like to simply use the payment formula in this situation, it turns out that we cannot. Instead, we need to use a different formula: one that will look very similar, but which will have important differences that compensate for the apparently “backwards” nature of the problem. The formula is this: A = Payment · 1+
APR nY n APR n
−1
where Variable Meaning A The amount of money saved after Y years Y The number of years over which payments are being made APR The annual interest rate n The number of compounding periods per year Payment The number of dollars saved in each compounding period We can use this formula to answer the question that we started with. We compute 180 · 1+
0.0225 12·5 12 0.0225 12
−1
0.11895438 0.001875 = 11, 419.62. = 180 ·
Then after 5 years, Mary will have $11, 419.62 in her account. Note that if she had simply kept this money in a piggy bank, she would only have $10, 800.
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TIM CARSTENS
We can use this formula to answer a related question: Problem. Mary wants to save up to buy a rare electric guitar, just like in Wayne’s World. The guitar costs $10, 000. Every month she deposits $200 into a savings account that earns 2.25% interest, compounded monthly. How long will she need to keep saving before she has enough money to buy the guitar? In this case, we setup the savings formula with all of the information we are given, and recognize that we need to solve for Y : 10, 000 = 200 · 1+
0.0225 12Y 12 0.0225 12
−1
.
This is a perfect situation to use logarithms. I will first simplify the numbers in the equation, and then will proceed to solve the problem using some algebra and logarithms. First I need to simplify the numbers: (1) (2) (3) (4) 1.00187512Y − 1 0.001875 1.00187512Y − 1 50 = 0.001875 0.09375 = 1.00187512Y − 1 10, 000 = 200 · 1.09375 = 1.00187512Y .
In the above, all I’ve done is use a calculator to collapse the numbers. I can now compute the logarithm of both sides of the equation. Remember that log(ab ) = b · log a, which allows me to write the following: (5) (6) (7) (8) (9) log 1.09375 log 1.09375 log 1.09375 log 1.001875 log 1.09375 12 · log 1.001875 3.98649528 = log 1.00187512Y = 12Y · log 1.001875 = 12Y = Y = Y.
So evidently she will need to save for roughly 4 years before she can buy the guitar. This same strategy would have worked with any numbers: the critical part was getting to line (4), so that we could introduce a logarithm on line (5), and then apply a basic property of logarithms on line (6). After that, we can use our calculator to perform the necessary computations.