Exponential and Logarithmic Models
Students compare exponential growth and decay models and use logarithms to solve equations arising from real-world situations.
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Exponential Growth and Decay
An exponential model has the form y = a · b^t, where equal time intervals multiply the quantity by the same factor b. If b > 1, the model shows growth; if 0 < b < 1, it shows decay. Percent models can be written y = a(1 + r)^t for growth and y = a(1 − r)^t for decay, with r expressed as a decimal. Suppose two quantities both begin at 500. An investment growing 6 percent per year is G(t) = 500(1.06)^t, while equipment losing 20 percent of its value per year is D(t) = 500(0.80)^t. After three years, G(3) is about 595.51, but D(3) is 256. Growth rises increasingly quickly, while decay falls quickly at first and then approaches zero without reaching it in the model.
Parameters and Initial Values
In y = a · b^t, the parameters a and b determine the model’s behavior. Setting t = 0 gives y = a because b^0 = 1, so a is the initial value. The base b is the growth or decay factor for one time interval. It equals 1 + r for growth or 1 − r for decay. For example, a car worth $28,000 that loses 18 percent of its value each year can be modeled by V(t) = 28,000(0.82)^t. Here, 28,000 is the initial value and 0.82 means the car retains 82 percent of its value each year. After three years, V(3) is about $15,438. Changing a vertically scales the entire graph, while changing b affects the direction and rate of change. The units of t must match the time interval represented by b.
Logarithms as Inverses
A logarithm answers the question, “What exponent produces this value?” The statement log_b(x) = y means exactly the same thing as b^y = x. For real-number logarithms, b must be positive and not equal to 1, and x must be positive. For example, because 2^5 = 32, log_2(32) = 5. Logarithms are useful when an unknown appears in an exponent. If an investment follows A(t) = 1,000(1.07)^t, the time required to double satisfies 1.07^t = 2. Therefore, t = log_1.07(2). A calculator may evaluate this using the change-of-base formula: log_1.07(2) = ln(2) ÷ ln(1.07), which is about 10.2 years. Exponential and logarithmic functions are inverses, so their graphs are reflections across the line y = x.
Solving Exponential Equations
To solve an exponential equation, first isolate the exponential expression and then apply a logarithm to both sides. Suppose a bacterial culture is modeled by P(t) = 800(1.35)^t, and scientists want to know when the population will reach 5,000. Set 800(1.35)^t = 5,000 and divide by 800 to obtain 1.35^t = 6.25. Taking natural logarithms gives ln(1.35^t) = ln(6.25). Use the power property of logarithms to rewrite the left side as t ln(1.35). Then t = ln(6.25) ÷ ln(1.35), which is approximately 6.11. Thus, the model reaches 5,000 bacteria after about 6.11 time units. Common logarithms could also be used because dividing log(6.25) by log(1.35) produces the same result. Technology is especially helpful when the logarithms are not simple whole numbers.
Interpreting Model Results
A calculated logarithmic solution must be interpreted using the situation’s units, domain, and practical constraints. Suppose the amount of a medication in the body is M(h) = 120(0.78)^h milligrams, where h is hours after a dose. To find when the amount falls to 30 milligrams, solve 120(0.78)^h = 30. This gives 0.78^h = 0.25 and h = ln(0.25) ÷ ln(0.78), or about 5.55 hours. The decimal is meaningful if time is measured continuously. If a nurse checks only at whole-hour intervals, the first reading expected to be below 30 milligrams occurs at 6 hours, when the model predicts about 27 milligrams. The result should also be checked against the graph and the reasonable domain h ≥ 0. A model predicts behavior under its assumptions; it does not guarantee an exact measurement for every patient.