Hello, a car was purchased for $20,000.
This is the initial value.
The car depreciates by 22% of each year.
After 1 year, the value is the initial value 20,000 minus 22% of 20,000.
[tex]20000-20\%*20000=20000\cdot (1-20\%)=20000\cdot (1-0.20)=20000\cdot 0.8=16000[/tex]
After 2 years, the value is.
[tex]20000\cdot 0.8\cdot 0.8=20000\cdot0.8^2=12800[/tex]
Let's take n a positive integer, after n years, the value is.
[tex]\large \boxed{\sf \bf \ 20000\cdot0.8^n \ }[/tex]
a) After 12 years, the value is.
[tex]20000\cdot0.8^{12}=1374.389...[/tex]
This is rounded to $1,374
b) We need to find n such that
[tex]20000\cdot0.8^n=100\\\\ln(20000)+nln(0.8)=ln(100)\\\\n=\dfrac{ln(100)-ln(20000)}{ln(0.8)}=23.74...[/tex]
This is around 23.75 meaning 23 years and 75% of 1 year (meaning 9 months).
So to be worth less than $100, 23 years and 9 months are required.
Thank you
