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A used car depreciates at a rate of f(x)=15500(0.68)^x. When will the car be worth $2,000?

1 Answer

8 votes

Answer:

5.31 years

Explanation:

Set
f(x)=2000 and solve for x (assuming x to be years):


f(x)=15500(0.68)^x\\\\2000=15500(0.68)^x\\\\(2000)/(15500)=0.68^x\\ \\(4)/(31)=0.68^x\\ \\ln((4)/(31))=ln(0.68^x)\\ \\ln((4)/(31))=xln(0.68)\\ \\x=(ln((4)/(31)))/(ln(0.68))\\ \\x\approx5.31

Therefore, the car will be worth $2000 after 5.31 years

User Mccc
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