Question

Is \(y=1.02^{t}\) an exponential decay model? Explain.

Short answer

No, \(y=1.02^{t}\) is not an exponential decay model since the base of its exponent (1.02) is greater than 1.

Step-by-step solution

Recognizing Exponential Decay

An exponential decay model is generally in the form \(y = a(1-r)^{t}\) or \(y = ab^{-t}\) where \(a\) is the initial value, \(r\) is the rate of decay, \(b\) is a positive number greater than 1, and \(t\) is time. For both forms, the base of the exponent (which effectively is the factor by which the quantity diminishes over each time period) is less than 1.

Comparing the Model to the General Form

In the equation \(y=1.02^{t}\), comparing to the general forms described in step 1, the base of the exponent is 1.02, which is greater than 1.

Making a Conclusion

Therefore, since the base of the exponent in \(y=1.02^{t}\) is greater than 1, it does not meet the requirement for an exponential decay model (which would have a base of the exponent less than 1).