4.3. Exponential and Logarithmic Models
Exponential and Logarithmic Models
While we have explored some basic applications of exponential and logarithmic functions, in this section we explore some important applications in more depth.
More complex exponential equations can often be solved in more than one way. In the following example, we will solve the same problem in two ways – one using logarithm properties, and the other using exponential properties.
Example 4.3.1a
In 2008, the population of Kenya was approximately 38.8 million, and was growing by 2.64% each year, while the population of Sudan was approximately 41.3 million and growing by 2.24% each year. If these trends continue, when will the population of Kenya match that of Sudan?
We start by writing an equation for each population in terms of t, the number of years after 2008.
To find when the populations will be equal, we can set the equations equal:
For our first approach, we take the log of both sides of the equation:
Utilizing the sum property of logs, we can rewrite each side:
Then utilizing the exponent property, we can pull the variables out of the exponent:
Moving all the terms involving t to one side of the equation and the rest of the terms to the other side:
Factoring out the t on the left:
Dividing to solve for t:
years until the populations will be equal.
Example 4.3.1b
Solve the problem above by rewriting before taking the log.
Starting at the equation:
38.8(1.0264)t = 41.3(1.0224)t
Divide to move the exponential terms to one side of the equation and the constants to the other side:
Using exponent rules to group on the left:
Taking the log of both sides:
Utilizing the exponent property on the left:
Dividing gives:
years
While the answer does not immediately appear identical to that produced using the previous method, note that by using the difference property of logs, the answer could be rewritten:
While both methods work equally well, it often requires fewer steps to utilize algebra before taking logs, rather than relying solely on log properties.
Radioactive Decay
In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes change over time. One of the common terms associated with radioactive decay is half-life.
Given the basic exponential growth/decay equation h(t) = abt , half-life can be found by solving for when half the original amount remains; by solving , or more simply . Notice how the initial amount is irrelevant when solving for half-life.
Example 4.3.2
Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?
We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant a. Then the decay of Bismuth-210 can be described by the equation Q(d) = a(0.87)d .
To find the half-life, we want to determine when the remainin