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A person invests 6000 dollars in a bank. The bank pays 5.75% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 11600 dollars?

Solution

Solution

Time : 11.47

Solution steps

Translate the problem into an equation:

11600 = 6000 (1 + \frac{0.0575}{365})^{365 t}

Solve for

t, 11600 = 6000 (1 + \frac{0.0575}{365})^{365 t} : t = 11.47

Time : 11.47

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