Question

The table below show the data to find an exponential model.

x 1 4 7 8 10

y. 798 1078 1519 2075 3102

1 goes with 798, 4 goes with 1078, 7 goes with 1519, 8 goes with 2075, and 10 goes with 3102

a. Write the data set to be used in order to linearize the data.


b. Write the system of equations


c. Write the matrices.


d. Indicate the c values


e. Write the exponential model in the form Y=C* 10^bx

Answer 1

a. The exponential model is of the form Y=Ce^bx. Taking natural logarithms of both sides of the equation, we get ln(Y)=ln(C)+bx. This form is linear in ln(Y).

The data set to be used to linearize the data is:

x ln(y)
1 ln(798)
4 ln(1078)
7 ln(1519)
8 ln(2075)
10 ln(3102)

b. The system of equations is:

$\begin{pmatrix}5 & \sum_{i=1}^{5} x_i \\ \sum_{i=1}^{5} x_i & \sum_{i=1}^{5} x_i^2 \\\end{pmatrix}\begin{pmatrix}b \\ ln(C) \\\end{pmatrix}=\begin{pmatrix}\sum_{i=1}^{5} ln(y_i) \\ \sum_{i=1}^{5} x_i ln(y_i) \\\end{pmatrix}$

c. The matrices are:

$\begin{pmatrix}5 & \sum_{i=1}^{5} x_i \\ \sum_{i=1}^{5} x_i & \sum_{i=1}^{5} x_i^2 \\\end{pmatrix}\begin{pmatrix}b \\ ln(C) \\\end{pmatrix}=\begin{pmatrix}\sum_{i=1}^{5} ln(y_i) \\ \sum_{i=1}^{5} x_i ln(y_i) \\\end{pmatrix}$

$\begin{pmatrix}5 & 30 \\ 30 & 266 \\\end{pmatrix}\begin{pmatrix}b \\ ln(C) \\\end{pmatrix}=\begin{pmatrix}7.986 \\ 27.214 \\\end{pmatrix}$

d. Solving for b and ln(C) using a calculator or a matrix-solving tool, we get:

b = 0.2695
ln(C) = 6.6195

Taking exponential of ln(C) we get C = 760.9305.

Therefore, the values of c are: C = 760.9305.

e. The exponential model in the form Y=C * 10^(bx) is:

Y = 760.9305 * 10^(0.2695x).