Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
7. Modeling with Exponential and Logarithmic Functions
Continue to next subchapter

Exercise 35 Page 348

Practice makes perfect
a We will first create a table of data pairs (x,ln y) using the given table.
Age, x 1 2 3 4 5
Weight, y 0.751 1.079 1.702 2.198 3.438
ln y -0.29 0.08 0.53 0.79 1.23

Next, we will plot the transformed points.

The points lie close to a line, so an exponential model should be a good fit for the original data. To write an exponential model y=ab^x, let's use the points ( 1, -29) and ( 5, 1.23) because they appear to be on the line. With this, we will write an equation in point-slope form. ln y- ln y_1= m(x- x_1) In this form, m is the slope and ( x_1, ln y_1) is a reference point that is on the line. Using the Slope Formula, we can substitute the chosen points and find the slope.
m = y_2-y_1/x_2-x_1
m = 1.23-( -0.29 )/5- 1
m=1.23+0.29/5-1
m=1.52/4
m=0.38
Now that we found the slope, we can write the equation in point-slope form. Let the point ( 1, -0.29) be our reference point. ln y+0.29=0.38(x-1) Next, we will isolate y to have the model in the form of y=ab^x.
ln y+0.29=0.38(x-1)
â–Ľ
Solve for y
ln y+0.29=0.38x-0.38
ln y=0.38x-0.67

e^(LHS)=e^(RHS)

e^(ln y)=e^(0.38x-0.67)

a = e^(ln(a))

y=e^(0.38x-0.67)
y=e^(0.38x)/e^(0.67)
y=e^(0.38)^x/e^(0.67)
y=(1.46228...)^x/1.95423...
y=1/1.95423...(1.46228...)^x
y=0.51170...(1.46228...)^x
y=0.51(1.46)^x
As a result, one possible model is y=0.51(1.46)^x depending on the choice of the points.
b We will first analyze the exponential growth model.

Exponential Growth Model y= a(1+ r)^t In the model, a is the initial amount and r is the percent increase written as a decimal. Let's rewrite the model in Part A in the form of y=a(1+r)^x y= 0.51(1+ 0.46)^t Therefore, we can identify the percent increase as 46 %. The weight of an Atlantic cod increases by 46 % each year.