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A person invests 3500 dollars in a bank. The bank pays 4.5% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 6900 dollars?

Solution

Solution

Time : 15.11

Solution steps

Translate the problem into an equation:

6900 = 3500 (1 + \frac{0.045}{12})^{12 t}

Solve for

t, 6900 = 3500 (1 + \frac{0.045}{12})^{12 t} : t = 15.11

Time : 15.11

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