Answer:
The price of car will be $1,900 after 14.9 years.
Step-by-step explanation:
Formula of depreciate:
![A=P(1-r)^n](https://tex.z-dn.net/?f=A%3DP%281-r%29%5En)
A= The price of the car after n years.
P= The initial price of the car
r= rate of depreciate
n=time in years.
Given that,
A new car is purchased for $ 15,100.
The value of car depreciates at 13% per year.
Here P=$15,100, A=$1,900, r=13%=0.13, n=?
![A=P(1-r)^n](https://tex.z-dn.net/?f=A%3DP%281-r%29%5En)
![\Rightarrow 1,900=15,100(1-0.13)^n](https://tex.z-dn.net/?f=%5CRightarrow%201%2C900%3D15%2C100%281-0.13%29%5En)
![\Rightarrow \frac{1,900}{15,100}=(0.87)^n](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%2C900%7D%7B15%2C100%7D%3D%280.87%29%5En)
![\Rightarrow (0.87)^n= \frac{1,900}{15,100}](https://tex.z-dn.net/?f=%5CRightarrow%20%280.87%29%5En%3D%20%5Cfrac%7B1%2C900%7D%7B15%2C100%7D)
![\Rightarrow (0.87)^n= \frac{19}{151}](https://tex.z-dn.net/?f=%5CRightarrow%20%280.87%29%5En%3D%20%5Cfrac%7B19%7D%7B151%7D)
Tanking ln function both sides
![\Rightarrow ln (0.87)^n= ln|\frac{19}{151}|](https://tex.z-dn.net/?f=%5CRightarrow%20ln%20%280.87%29%5En%3D%20ln%7C%5Cfrac%7B19%7D%7B151%7D%7C)
![\Rightarrow nln (0.87)= ln|\frac{19}{151}|](https://tex.z-dn.net/?f=%5CRightarrow%20nln%20%280.87%29%3D%20ln%7C%5Cfrac%7B19%7D%7B151%7D%7C)
![\Rightarrow n= \frac{ln|\frac{19}{151}|}{ln (0.87)}](https://tex.z-dn.net/?f=%5CRightarrow%20n%3D%20%5Cfrac%7Bln%7C%5Cfrac%7B19%7D%7B151%7D%7C%7D%7Bln%20%280.87%29%7D)
⇒n ≈14.9
The price of car will be $1,900 after 14.9 years.