Chapter 3, Section 1 Notes & Review

Welcome to Exponential Functions

Exponents

To start, let's evaluate these exponents (use a calculator if you please):

a.)5^2.6

b.)2^(4/5)

c.)3^(-3.2)

Pretty easy, right? All you need to do is type it into your calculator.

Its just important that you know how to use an exponent.

Interest

There are two kinds of interest: compound and continuous. The formulas are as follows:

Compound Interest:  A=P(1+r/n)^nt

Continuous Interest: A=Pe^rt

Here's what that means: A: amount, or the total money you now have

                                        P: principle, or the amount you put into the bank

                                        R: rate, or the rate (as the decimal)

                                        N: Number, or number times compounded per year (compound only)  

                                        T: Time, in years.  Also known as Mr.

                                        e is a "natural base," equal to about 2.71828.  It is a number, not a variable, it doesn't change based                                    on the problem. It's used for continuous interest (its used for logs and other things later on; but                                        fret not young padawan, this is not the time)

Tips: A formula is continuous if is says it is.

         Sometimes a problem will say it is compounded continuously. Do not be fooled! If an interest rate is continuous, that means it is compounded continuously.

          If it says annually, monthly, quarterly, or any division of a year, then it is not compounded conitinuously, just compounded.

          There is a natural log key on your graphing calculator- far right, five up. Hit second, then the division key.

Let's try it!!!!!!

Now we just need to come up with some problems:

Invest  $7600 at a rate of 5.6%, compound annually.  Find the balance after 4 years.

We need to find A. The first step is: what variables do we have?  Well first off, the rate is given as 5.6%.  Divide this by 100 to receive the proper decimal for the equation.  Also, the intial investment is given ($7600), this number will be the variable P.  We must find the balance after 4 years, so 4 would be the variable t.  Since the investment is compounded annually, set N = 1.

Here's where we are right now:

A = ?               r = .056               n = 1               P = 7600               t = 4

Now all you have to do is plug the variables into the given equation.  Since this is compound interest, use the A=P(1 + r/n)^nt equation.

A = 7600 (1 + (.056/1))^(1 * 4)

After using a calculator to solve, A comes out to be: $9450.82

Lets Try it!!! Again!!! Continuously!!!

If you are paid $9,000 after 3 years with 6.3% interest, how much did you invest?

 We're compounding continuously this time, so the formula we'll use is A=Pe^rt

What do we know, and what do we need?

A=9,000      r=.063            t=4         e=2.71828       P=?

These are actually easier, there are just more decimals. Also, keep in mind that in this is particular problem we're not asking for the total, you have that already. We're asking for how much you put in. Now that you have all but one variable just solve it like an algebra problem.

9,000=Pe^(.063)(4)

9,000=P1.29

9,000/1.29=P

P=6995.2

So if you want 9,000 dollars 4 years from now, put $6995.20 in a bank account compounded continuously at 6.3% interest.

Tip: Remember, you know what e is. It is not a variable.