Chapter 7.2, Problem 15PPS

(a)

To determine

To find: the exponential function for the given condition.

Answer to Problem 15PPS

The exponential function for the given condition is $y=10,000{\left(1.045\right)}^n$ .

Explanation of Solution

Given information:

The given condition is given by “My-Lien received $\$10,000$ from her grandmother. Her parents invested all of the money, and by 2015, the amount will have grown to $\$16,960$ ”.

The general model can be written as $I_0{\left(1+i\right)}^n=V$ and the general exponential function is $y=a{b}^x$ .

Substitute 12 for $n$ , $10,000$ for $I_0$ , and $16,960$ for $V$ in the function $I_0{\left(1+i\right)}^n=V$ to find the value of $i$ .

  $\begin{array}{c}10,000{\left(1+i\right)}^{12}=16,960\\ {\left(1+i\right)}^{12}=\frac{16,960}{10,000}\\ i={\left(1.696\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}-1\\ =4.5\%\end{array}$

Substitute $10,000$ for $I_0$ , and $4.5\%$ for $i$ in the function $I_0{\left(1+i\right)}^n=y$ to find the exponential function.

  $\begin{array}{c}y=10,000{\left(1+0.045\right)}^n\\ =10,000{\left(1.045\right)}^n\end{array}$

Therefore, the exponential function for the given condition is $y=10,000{\left(1.045\right)}^n$ .

(b)

To determine

To find: the balance in the account in the year 2005.

Answer to Problem 15PPS

The balance in the account in the year 2005 is $\$26,337$ .

Explanation of Solution

Given information:

The given condition is given by “My-Lien received $\$10,000$ from her grandmother. Her parents invested all of the money, and by 2015, the amount will have grown to $\$16,960$ ”.

Substitute $22$ for $n$ in the function $y=10,000{\left(1.045\right)}^n$ to find the exponential function.

  $\begin{array}{c}y=10,000{\left(1.045\right)}^{22}\\ =10,000\left(2.6337\right)\\ =26,337\end{array}$

Therefore, the balance in the account in the year 2005 is $\$26,337$ .