Answer:
D) 9 years.
Step-by-step explanation:
We have been given that Gary bought a car for $40,000 and equation \(V=40,000(0.85)^t\) represents the value of the car after t years.
First of all we will find the one-fourth of 40,000.
\(\text{One-forth of car's purchase price}=\frac{\$40,000}{4}\)
\(\text{One-forth of car's purchase price}=\$10,000\)
To find the time it will take the car to be worth less than one-fourth of its purchase price, we will substitute V=10,000 in our given equation.
\(10,000=40,000(0.85)^t\)
Let us divide both sides of our equation by 40,000.
\(\frac{10,000}{40,000}=\frac{40,000(0.85)^t}{40,000}\)
\(0.25=0.85^t\)
Let us take natural log of both sides of our equation.
\(ln(0.25)=ln(0.85^t)\)
Using natural log property \(ln(a^b)=b*ln(a)\) we will get,
\(ln(0.25)=t*ln(0.85)\)
\(\frac{ln(0.25)}{ln(0.85)}=\frac{t*ln(0.85)}{ln(0.85)}\)
\(\frac{-1.3862943611198906}{-0.1625189294977749}=t\)
\(8.530048563597=t\)
Upon rounding our answer to the nearest year we will get,
\(t\approx 9\)
Therefore, it will take 9 years the car to be worth less than one-fourth of its purchase price and option D is the correct choice.