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# Use of Exponential Growth Functions

Solution to Question 2

For continuous compound interest, , where P = initial amount invested or borrowed, n = number of periods of compounding, i = interest rate for every period, and A = future amount of investment after n periods. Given that A = \$50,000, n = 17 years (because P should grow to \$50,000) at beginning of 18th year), and i = 3.5% or 0.035, then:

Therefore, Jack and Susan needs to deposit \$27,859.81 now to ensure that their daughter will afford college fees of \$50,000 when she turns 18 years old if the deposit will be compounded continuously at an annual rate of 3.5%.

Context

This problem relates to exponential functions, which model real world issues involving original values that increase or decrease exponentially. Mathematically, these functions are represented by  where b = the original value or base, x = any real number, and y = value of b to power x. Compound interest equation, represented by   is one of the different applications of exponential functions that model initial values that grow exponentially. In this case, the initial amount compounded continuously increases exponentially because the interest rate applies to the cumulative sumo of the original amount and yearly interest on the amount.

Method

The compound interest problem could have been solved through three different methods: stepwise approach, use of table of compound interests, and use of compound interest mathematical model. In the stepwise approach, the problem could have be solved by using simple interest formula ( to calculate cumulative sum of initial deposit and interest on the deposit every year. However, this method would require the interest earned at the 17th year to be known to facilitate calculation of the cumulative sum for the preceding years. Therefore, it was not possible to use the method because the interest was not given.

In use of table of compound interests, the future value of \$1 could have been read from the table as the value corresponding to cell at the 17th year row and 3.5% column. The quotient of the \$50,000 and this value would have given the initial amount to be deposited. However, this method was not used because the tables were not available.

The use of compound interest model was selected to solve the problem because it was easier to use. All variables required in the formula were provided. In addition, it only needed substitution of the given values to corresponding variables in the formula and working out to obtain the unknown variable, that is, initial amount to be invested.

Explanation

The problem was solved using the formula of compound interest, given by  The given values were substituted to the corresponding variables to facilitate calculation of the amount to be deposited (P) by Jack and Susan. After substitution, the values inside the bracket were added to eliminate the expression. This followed calculating the addition results within the brackets to power 18 in order to eliminate the brackets. The result from this step was multiplied by the unknown variable (P) to do away with expression on the right hand side of the equation. Finally, both sides of the equation were divided by the constant value at the right hand side (1.7947) to obtain value of P (\$27,859.81), which was taken to be the deposit that will grow exponentially to \$50,000 by the time the daughter of Jack and Susan turns 18 years old.

Conclusion

Using the equation of compound interest, Jack and Susan needs to invest \$27, 859.81 today to ensure that their daughter will have \$50,000 for college at 18 years. This amount comprises of cumulative amount and interest on this amount at the end of the 17th year. This amount is correct because the girl will start the college at the start of the 18th year.