Respuesta :
We have two cases in which we need to write the formula for exponential growth or exponential decay, and we have the following two cases:
• A population of 300 decreases by 8% each year.
,• A $27000 car depreciates 14.5% each year.
1. We can see that we have two cases of exponential decay, and the general formula in both cases is as follows:
[tex]\begin{gathered} y=a(1-r)^x \\ \\ \text{ Which is equivalent to:} \\ \\ A(t)=a(1-r)^t \end{gathered}[/tex]Where:
• a is the initial value (that is, the amount before starting the decay).
,• r is the decay rate (the decay rate is given as a percentage, and we can express it as a decimal)
,• t is the number of time intervals that have passed
2. And we can analyze both cases as follows:
First Case: A population of 300 decreases by 8% each year.
• The initial value, a = 300 (the initial population)
• The decay rate is 8% or we can express it as follows:
[tex]r=8\%=\frac{8}{100}=0.08[/tex]Now, we can express the scenario as follows:
[tex]\begin{gathered} A(t)=300(1-0.08)^t=300(0.92)^t \\ \\ A(t)=300(0.92)^t \end{gathered}[/tex]Second Case: A $27000 car depreciates 14.5% each year
We can proceed as before, and we have:
• a = $27000
• r = 14.5% which is equivalent to:
[tex]\begin{gathered} r=14.5\%=\frac{14.5}{100}=0.145 \\ \\ r=0.145 \end{gathered}[/tex]Therefore, we can express the situation as follows:
[tex]\begin{gathered} A(t)=27000(1-0.145)^t \\ \\ A(t)=27000(0.855)^t \\ \end{gathered}[/tex]Therefore, in summary, we can express both situations as follows:
• A population of 300 decreases by 8% each year:
[tex]A(t)=300(1-0.08)^{t}=300(0.92)^{t}[/tex]• A $27000 car depreciates 14.5% each year:
[tex]A(t)=27000(1-0.145)^t=27000(0.855)^t[/tex]