Respuesta :
Answer:
In 2026 car will have a value of $5,000.
Step-by-step explanation:
We have been given that Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year.
Since we know that an exponential function is in form: [tex]y=a*b^x[/tex], where,
a = Initial value,
b = For decay or decrease b is in form (1-r), where r represents decay rate in decimal form.
Let us convert our given decay rate in decimal form.
[tex]14\%=\frac{14}{100}=0.14[/tex]
Upon substituting a =28,000 and r=0.14 in exponential decay function we will get,
[tex]y=28,000(1-0.14)^x[/tex], where x represents number of years after 2014.
Therefore, the function [tex]y=28,000(0.86)^x[/tex] represents the value of car x years after 2014.
To find the number of years it will take to car have the value of $5,000, we will substitute y=5,000 in our function.
[tex]5,000=28,000(0.86)^x[/tex]
Let us divide both sides of our equation by 28,000.
[tex]\frac{5,000}{28,000}=\frac{28,000(0.86)^x}{28,000}[/tex]
[tex]0.1785714285714286=(0.86)^x[/tex]
Let us take natural log of both sides of our equation.
[tex]ln(0.1785714285714286)=ln((0.86)^x)[/tex]
Using natural log property [tex]ln(a^b)=b*ln(a)[/tex] we will get,
[tex]ln(0.1785714285714286)=x*ln(0.86)[/tex]
[tex]\frac{ln(0.1785714285714286)}{ln(0.86)}=\frac{x*ln(0.86)}{ln(0.86)}[/tex]
[tex]\frac{-1.7227665977411033893}{-0.1508228897345836}=x[/tex]
[tex]x=11.422447\approx 12[/tex]
As in the 12th year after 2014 car will have a value of $5,000, so we will add 12 to 2014 to find the year.
[tex]2014+12=2026[/tex]
Therefore, in 2026 car will have a value of $5,000.
Answer: 1) [tex]28000(0.86)^x[/tex]
2) After 11.422 years( approx ) since 2014 the value of car will be 5000.
Step-by-step explanation:
Since, the initial value of the car = 28000
The yearly decrease in the value of car = 14 %
Hence, the value of car after x years
= [tex]28000(1-\frac{14}{100})^x[/tex]
= [tex]28000(1-0.14)^x[/tex]
= [tex]28000(0.86)^x[/tex]
Which is the required exponential function that shows the value of car.
Now let after t years, the value of car is 5000,
[tex]\implies 28000(0.86)^y=5000[/tex]
[tex]\implies 28(0.86)^y=5[/tex]
[tex]\implies (0.86)^y = 0.178571429[/tex]
By taking log both sides,
[tex]\implies y log(0.86)=log(0.178571429)[/tex]
[tex]\implies y=11.4224479\approx 11.422\text{ years}[/tex]
