Exercise 49 Page 330

Writing the Equation

Our starting deposit is $500, and each year we gain 6 % of the previous value. Let's form an equation where y represents the amount of money in the account and t represents years. Since the amount in the account increases by the same factor each year, we will use the general form of an exponential growth function. y=a(1+r)^t In the 0^(th) year, there should be $500 in the account. This tells us that when t=0, y=500. 500=a(1+r)^0 ⇒ 500=a

Let's substitute this for the a-value in our equation. y=500(1+r)^x Each year the account gains 6 %. This can be rewritten as 0.06. In our exponential growth function, r represents the rate of growth, so we will substitute 0.06 for r. y=500(1+0.06)^x ⇒ y=500(1.06)^x

Finding When $800 are in the Account

Now that we have our equation, we can find when the balance will be $800. We will substitute 800 for y and solve for x. 800=500(1.06)^x It will be difficult to find like bases for this equation, so we will solve this by graphing. We will write our equation as a system of equation in our calculator by pressing the Y= button and typing the functions in the rows.

Resize the window by pushing WINDOW and changing the settings.

To find the point of intersection, push 2nd and CALC and choose the fifth option, intersect. Choose the first and second curve, and pick a best guess for the point of intersection.

As we can see in the bottom of the graph window, the equation has a solution when x is approximately 8.09. There will be $800 after 8.09 years.

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Writing the Equation

Finding When $800 are in the Account