Respuesta :
Given the model:
[tex]y=250(1.55)^t[/tex]The model above represents the value of a pair of OFF-white Jordan, where t is the number of years since the shoe was released.
Let's answer the following questions.
• a) Let's determine if the model represents an exponential growth or decay.
Apply the exponential function:
[tex]y=a(b)^x[/tex]When b is greater than 1, we can say the function reppresents exponential grwoth.
When b is less than 1, the funcrion represents exponential decay.
Here, b = 1.55 is greater than 1.
Since b is greater than 1, the given model represents an exponential growth.
• b) Take the exponential growth function:
[tex]y=a(1+r)^x[/tex]Where:
r = percentage increase.
Since the model is an exponential growth, we are to find the annual percent increase.
To find the annula percentage increase, we have:
[tex]\begin{gathered} y=250(1.55)^t \\ y=250(1+(1.55-1)^t \\ \\ \\ y=250(1+0.55)^t \\ \\ r=0.55\Longrightarrow\text{ 55\%} \end{gathered}[/tex]Therefore, the annual percent increase (r) is = 55%
• c) To find when the resale value of the shoe will be over 2000, substitute 2000 for y and solve for t.
Thus, we have:
[tex]\begin{gathered} y=250(1.55)^t \\ \\ 2000<250(1.55)^t \end{gathered}[/tex]Let's solve for t.
Divide both sides by 250
[tex]\begin{gathered} \frac{2000}{250}<\frac{250(1.55)^t}{250}^{} \\ \\ 8<1.55^t \end{gathered}[/tex]Take the natural logarithm of both sides:
[tex]\begin{gathered} \ln 8Divide both sides by ln1.55:[tex]\begin{gathered} \frac{\ln 8}{\ln 1.55}<\frac{t\ln1.55}{\ln1.55} \\ \\ \frac{\ln 8}{\ln 1.55}\frac{\ln8}{\ln1.55}=\frac{2.079}{0.438} \\ \\ t>4.7\approx5 \\ \\ \end{gathered}[/tex]Therefore, the shoe's resale value will be over $2000 after 5 years
ANSWER:
• a) Exponential growth
,• b) 55%
,• c) 5 years
