APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(Population Word Problems)

To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation. In this section, we will review population problems.

Example 16: Convert the exponential equation

to an equivalent equation with A) base e, B) base 10, and C) base 23.

Solution and Explanations:

Let's make a few observations before we begin. Since the base is 1.12, greater than than 1, the value of f(t) will get larger as t gets larger. We are going to find a few points that satisfy this equation and use these points to find the new equations.

When t = 0, the value of f(t) is 5,000, and the corresponding point is (0, 50,000).

At t = 1, the value of f(t) is , and the corresponding point is (1, 5,600).

At t = 2, the value of f(t) is , and the corresponding point is (2, 6,272).

At t = 3, the value of f(t) is , and the corresponding point is (23, 7,024.54).

Base e:

• Let's start with a generic exponential equation with base e:

  

• Substitute the point (0, 5,000) in the equation.

  

  The equation can now be written

  

• Substitute the point (1, 5,600) into the equation

  

• Divide both sides of the equation by 5,000:

  

• Solve for b by taking the natural logarithm of both sides of the equation

  

• Simplify the right side of the equation using the third rule of logarithms:

  

• Simplify the left side of the equation:

  

  rounded to 0.11333.

• Insert this value of b in the above equation:

  

• Let's check the model (equation) with the point (2, 6,272):

  

  The model (equation) is correct.

The equation

is equivalent to the equation

Observation: , the original base.

Base 10:

• Let's start with a generic exponential equation with base 10:

  

• Substitute the point (0, 5,000) in the equation. . The equation can now be written

  

• Substitute the point (1, 5,600) into the equation :

  

• Divide both sides of the equation by 5,000:

  

• Solve for b by taking the common logarithm of both sides of the equation .

  

• Simplify the right side of the equation using the third rule of logarithms:

  

• Simplify the left side of the equation:

  

  rounded to 0.0492.

• Insert this value of b in the above equation:

  

• Let's check the model (equation) with the point (2, 6,272):

  

  This is close enough. Recall that it won't check exactly because we rounded the value of b.

  

The equation

is equivalent to

Observation: , the original base.

Base 23:

• Let's start with a generic exponential equation with base 23:

  

• Substitute the point (0, 5,000) in the equation.

  

  The equation can now be written

  

• Substitute the point (1, 5,600) into the equation

  

• Divide both sides of the equation by 5,000:

  

• Solve for b by taking the logarithm to the base 23 of both sides of the equation

  

• Simplify the right side of the equation using the third rule of logarithms:

  

• Simplify the left side of the equation:

  

• b = 0.036144

• Insert this value of b in the above equation:

  

• Let's check the model (equation) with the point (2, 6,272):

  

The equation

is equivalent to the equation

Observation: , the original base.

If you would like to work another example, click on Example

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