Compound Interest e - Department of Mathematics

Compound and Continuous Interest:

NOTE: Interest problems are scattered throughout chapter 4. I am doing them altogether in one worksheet here.

You will need to memorize the following two formulas:

r nt A=P 1+

n In both cases, the variables represent:

A = P ert

A: Final amount r: Growth rate (in decimal) P : Initial amount t: Time

The formula on the left is called Compound Interest, with the compounding factor n. The formula on the right is called continuous interest, which uses the exponential Euler number e. It will be up to you to look at the context clues of the word problem to determine which formula to use.

The effective rate or effective yield is the actual percentage increase in one year that you earn on an investment or pay on a loan after the effects of compounding frequency are considered. So using the formulas above, we have two effective rate/yield formulas:

rn ER = 1 + - 1

n

ER = er - 1

Just like the previous set of formulas, the formula on the left is the effective yield due to compound interest, while the one on the right is the effective yield due to continuous interest.

Examples: Answer the following questions. Round all your answers (for this worksheet) correct to two decimal places.

1. An investment of $2500 is put into a bank that has an annual rate of 6% compounded monthly. Find the principal after one month. Assuming that no money was added or withdrawn from the account, What will the balance be after 20 years?

2. If a savings fund pays interest at a rate of 10% per year compounded semiannually, how much money invested now will amount to $6500 after 5 years?

3. Approximate the effective yield corresponding to an interest rate of 5% per year compounded quarterly. Now do the same for compounded continuously.

4. Determine the nominal rate r if a an account compounding monthly results in an effective yield of 4.75%. Express your answer as a percent.

5. Determine the per annum interest rate r required for an investment with continuous compound interest to yield an effective rate of 5.25%. Express your answer as a percent.

6. How much money, invested at an interest rate of r% per year compounded continuously, will amount to A dollars after t years? A = 6000, r = 6.1, t = 14.

7. An investment is made at 7% per annum compounded continuously. Determine the time T in years required for the investment to double.

8. Determine the per annum interest rate r required for an investment with continuous compound interest to triple in 7.75 yrs.

9. Money is invested at an interest rate r compounded continuously. Express the time required for the money to quadruple as a function of r. Now express the rate required for the money to quadruple as a function of t.

Section 4.6: Exponential Equations

This worksheet brings together a lot of computational problems involving both exponential and logarithmic properties. If you haven't already done so, you need to memorize the Exponential Properties of Section 4.2 and the Logarithmic Properties of Section 4.5.

Example: Simplify the following equation completely:

e5x - e-5x 2 - e5x + e-5x 2

We have previously worked with equations where you have an exponential term on one side and a number on the right, like 53x = 7. In these cases, changing from exponential form to logarithm form cleans up your problem.

Now we will deal with equations that have exponential terms on either side of the equation. Solve for x for the following. The first two should be a review, while the last two will be new.

Example: Solve for x for the following equations: 1. 24x+4 = 8

3. 45x+12 = 63x

2. 85x = 1/16

4. 34-x = 52-x

The next set of equations deal with what's called exponential quadratics. They look like quadratic functions, but they have exponential terms incorporated in them. Again, be careful not to confuse coefficients with bases. To start the problems, you use a method called "u" substitution to transform your exponential equation into a quadratic equation. You then solve the quadratic. Don't forget to convert back at the end. Example: Use the substitution method to solve for x for the following equations:

1. e2x - 8ex - 15 = 0

2. ex - 24e-x = -5

3. 2x + 8 ? 2-x = 6

And finally, you have to work with computing inverses. Remember that the first step is to switch x and y. After that, you use mathematical properties to try to solve for y. The end result will be your inverse. Try the following:

1. Compute the inverse of the following function: 10x + 10-x

y= 8

2. Compute the inverse of the following function: ex + e-x

y= 12

3. Compute the inverse of the following function: e5x + e-5x

y = e5x - e-5x

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