Answer:
5.31 years
Step-by-step explanation:
Set [tex]f(x)=2000[/tex] and solve for x (assuming x to be years):
[tex]f(x)=15500(0.68)^x\\\\2000=15500(0.68)^x\\\\\frac{2000}{15500}=0.68^x\\ \\\frac{4}{31}=0.68^x\\ \\ln(\frac{4}{31})=ln(0.68^x)\\ \\ln(\frac{4}{31})=xln(0.68)\\ \\x=\frac{ln(\frac{4}{31})}{ln(0.68)}\\ \\x\approx5.31[/tex]
Therefore, the car will be worth $2000 after 5.31 years
