A deposit of $100 is made into an account earning j₁₂ = 18%.Another $100 is deposited into a 2^nd account earning j₂ = 10%.At what time,n (where n is in years),would the accumulated value of the first account be twice as much as the accumulated value in the 2^nd account? (Answer in years)

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Quizplus · Accepted Answer

The accumulated value of the first account is $100(1 + 0.18/12)^(12n) = $100(1.015)^12n.The accumulated value of the second account is $100(1 + 0.10/12)^(12n) = $100(1.00833)^12n.We want to find n such that $100(1.015)^12n = 2 * $100(1.00833)^12n.Dividing both sides by $100(1.00833)^12n, we get (1.015)^12n = 2.Taking the 12th root of both sides, we get 1.015^n = 2^(1/12).Solving for n, we get n = log(2^(1/12)) / log(1.015) = 9.3 years.

A Deposit of $100 Is Made into an Account Earning