Question 11 • Financial Algebra/Mathematics
Set the value equal to $1,000 to find when it reaches the threshold: 2800(0.82)^x = 1000. Solving gives x = ln(1000/2800) ÷ ln(0.82) ≈ 5.19 years. The value drops below $1,000 just after this point, so the matching choice is 5.19 years.
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