Module 6: Lesson 2

 

Self-Check 3
  1. “Your Turn” from “Example 3” on page 363 of the textbook

    To solve this question the compound interest formula will need to be used.

     

    A = 1490

     

    P = 1000

     



     

    n = ?

    Thus, substituting the known values into the compound interest formula gives the following:

     



    This question can be solved by using either a systematic trial or by graphing. A possible solution to the systematic trial is shown.

    Try n = 2. (1.0415)2 = 1.0847…, which is less than 1.49.

    Try n = 4. (1.0415)4 = 1.1766…, which is less than 1.49.

    Try n = 8. (1.0415)8 = 1.384 44…, which is less than 1.49.

    Try n = 12. (1.0415)12 = 1.628 96…, which is greater than 1.49. Thus, n is between 8 and 12.

    Try n = 10. (1.0415)10 = 1.5017…, which is greater than 1.49. Thus, n is between 8 and 10.

    Try n = 9. (1.0415)9 = 1.441 89…, which is less than 1.49. Thus, it will take 10 pay periods for $1000 to reach $1490. This means that (because there are 2 compound periods per year) it will take 5 yr to reach $1490.

    The solution that can be obtained by using graphing technology follows:

     

    This is a graph of one exponential function and a constant function. The exponential function is y equals 1000 times 1.0415 to the exponent x. The constant function is y equals 1490. The intersection point of the two functions is approximately (9.8071, 1490).

    Thus, the approximate solution is x = 9.8071. Because the number of pay periods must be a whole number, n = 10 pay periods. This means that (because there are 2 compound periods per year) it will take 5 yr to reach $1490.



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