Question

Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year, write an exponential model for the value of the car. Then estimate the year the car will have a value of 5,000

Answer 1

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: $y=a*b^x$, 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.

$14\%=\frac{14}{100}=0.14$

Upon substituting a =28,000 and r=0.14 in exponential decay function we will get,

$y=28,000(1-0.14)^x$, where x represents number of years after 2014.

Therefore, the function $y=28,000(0.86)^x$ 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.

$5,000=28,000(0.86)^x$

Let us divide both sides of our equation by 28,000.

$\frac{5,000}{28,000}=\frac{28,000(0.86)^x}{28,000}$

$0.1785714285714286=(0.86)^x$

Let us take natural log of both sides of our equation.

$ln(0.1785714285714286)=ln((0.86)^x)$

Using natural log property $ln(a^b)=b*ln(a)$ we will get,

$ln(0.1785714285714286)=x*ln(0.86)$

$\frac{ln(0.1785714285714286)}{ln(0.86)}=\frac{x*ln(0.86)}{ln(0.86)}$

$\frac{-1.7227665977411033893}{-0.1508228897345836}=x$

$x=11.422447\approx 12$  

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.

$2014+12=2026$

Therefore, in 2026 car will have a value of $5,000.

Answer 2

Answer: 1)  $28000(0.86)^x$

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

= $28000(1-\frac{14}{100})^x$

= $28000(1-0.14)^x$

= $28000(0.86)^x$

Which is the required exponential function that shows the value of car.

Now let after t years, the value of car is 5000,

$\implies 28000(0.86)^y=5000$

$\implies 28(0.86)^y=5$

$\implies (0.86)^y = 0.178571429$

By taking log both sides,

$\implies y log(0.86)=log(0.178571429)$

$\implies y=11.4224479\approx 11.422\text{ years}$