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If \$4000 is deposited at the end of each year in an account that earns 6.2% compounded semiannually, how long will it be before the account contains \$120,000?

when is 4000(1.031)^n = 12000 ?
1.031^n = 3
take log of both sides, and use log rules

n log 1.031 = log 3
n = log3/log1.031 = appr 35.99
So it would take 36 half-years or 18 years

check:
4000(1.031)^36 = 12005.27 , not bad

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