Question

use the properties of exponents to rewrite ( y = 5e^{-0.7t} ) in the form ( y = a(1 + r)^t ) or ( y = a(1 - r)^t ). round the value of ( r ) to the nearest thousandth. then find the percent rate of change to the nearest tenth of percent.
the rewritten function is ( y approx square ).
the percent decrease is about ( square % ).

Explanation:

Step1: Recall the formula \(e^{x}=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\). For small \(x\), \(e^{x}\approx1 + x\)

We have \(y = 5e^{- 0.7t}\). Using the approximation \(e^{x}\approx1 + x\) (when \(x=-0.7t\) and considering the form \(y=a(1 + r)^{t}\) or \(y=a(1 - r)^{t}\)), we know that \(e^{-0.7t}=(e^{-0.7})^{t}\).
Calculate \(e^{-0.7}\approx1-0.7+\frac{(-0.7)^{2}}{2}-\frac{(-0.7)^{3}}{6}+\frac{(-0.7)^{4}}{24}\)

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So \(y = 5e^{-0.7t}\approx5(0.479)^{t}\). Since \(0.479 = 1-0.521\), the function is in the form \(y=a(1 - r)^{t}\) where \(a = 5\) and \(r\approx0.521\)

Step2: Calculate the percent rate of change

The percent rate of change is given by \(r\times100\%\). Since \(r\approx0.521\), the percent rate of change is \(r\times100\%=52.1\%\)

Answer:

The rewritten function is \(y\approx5(1 - 0.521)^{t}\). The percent decrease is about \(52.1\%\)