APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

TIONS 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. We will also discuss why the base of e is used so often with population problems.

Example 1: Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. Write an equation with base 2 to determine the number (population) of cells after one hour.

Solution and Explanations:

First record your observations by making a table with two columns: one column for the time and one column for the number of cells. The number of cells (size of population) depends on the time. If you were to graph your findings, the points would be formed by (specific time, number of cells at the specific time). For example, at t = 0, there is 1 cell, and the corresponding point is (0, 1). At t = 1, there are 2 cells, and the corresponding point is (1, 2). At t = 2, there are 4 cells, and the corresponding point is (2, 4). At t = 3, there are 8 cells, and the corresponding point is (3, 8).

It appears that the relationship between the two parts of the point is exponential. At time 0, the number of cells is 1 or 2⁰ = 1. After 1 minute, when t = 1, there are two cells or 2¹ = 2. After 2 minutes, when t = 2, there are 4 cells or 2² = 4.

Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model)

f (t) = 2^t.

Determine the number of cells after one hour:

• Convert one hour to minutes.  ^.  = 60 min
  
  
  
• Substitute 60 for t in the equation. f (t) = 2^t:
  
  
  f (60) = 2⁶⁰ = 1.15×10¹⁸
  
  
  
  
  

Example 2: Determine how long it would take the population (number of cells) to reach 100,000 cells.

Solution and explanation:

• In this example, you know the number of cells at the beginning of the experiment (1) and at the end of the experiment (100,000), but you do not know the time. Substitute 100,000 for f(t) in the equation f (t) = 2^t:
  
  
  
  100, 000 = 2^t
• Take the natural logarithm of both sides:
  
  ln(100, 000) = ln(2^t)
  
  
  
• Simplify the right side of the equation using the third rule of logarithms:
  
  
  
  
  ln(100, 000) = t ln(2)
  
  
  
  
• Divide both sides by ln(2):
  
  t =  = 16.60964 min

It would take 16.6 minutes, rounded, for the population (number of cells) to reach 100,000.

Example 3: Write an equation with base 5 that is equivalent to the equation

f (t) = 2^t.

Solution and Explanation:

• Let's start with a generic exponential equation with base 5:
  
  
  
  
  
  
  
  
  f (t) = a ^. 5^bt.
  
  
  
• The f(t) represents the size of the population at time t, the t represents the time, and the a and b represent adjusters when we change the base. The value of a is the number of cells (size of population) at the beginning of the study, and the value of b is the relative growth rate based on a base of 5. We need to find the values of a and b.
  
• We know that the population is 1 at time 0, so insert these numbers in the equation
  
  f (t) = a ^. 5^bt.
  
  We have
  
  1 = a ^. 5^b . 0 = a ^. 5⁰ = a ^. 1 = a.
  
  We now know that the value of a in the adjusted equation is 1.
  
• Rewrite the equation
  
  f (t) = a ^. 5^bt
  
  with a = 1.
  
  f (t) = 1 ^. 5^bt
  
  which in turn can be rewritten as
  
  
  
  f (t) = 5^b . t.
• We know that the population after 1 minute is 2 cells, so insert these numbers in the equation
  
  f (t) = 5^b . t
  
  to obtain
  
  2 = 5^b . 1.
  
  
  
  
• Solve for b by taking the natural logarithm of both sides of the equation 2 = 5^b.
  
  ln(2) = ln(5^b).
  
  
  
• Simplify the right side of the equation using the third rule of logarithms:
  
  ln(2) = b ln(5).
  
  
  
• Divide both sides of the equation by ln(5) and simplify:
  
  b =  = 0.43067658075,
  
  
  rounded to 0.4307.
  
  
• Insert this value of b in the equation
  
  f (t) = 5^b . t,
  
  and the equation is simplified to
  
  f (t) = 5^0.4307t
  
  
  
• We know that the population is 8 after 3 seconds, so use these values to check the validity of the above equation. Substitute 3 for t in the right side of the above equation. If the answer is 8, or close to 8 because we rounded, then the model (equation) is correct. 5^0.4307(3) = 8.00906, rounded to 8.
  
• We know from the original equation that after 4 seconds, the population is 16. Let's do a second check. 5^0.4307(4) = 16.02415, rounded to 16 cells. The check would be closer had we rounded b to more decimals.