A person invests 6500 dollars in a bank. The bank pays 4% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 8900 dollars?

Solution

Solution

Time : 7.86

Solution steps

Translate the problem into an equation:

8900 = 6500 (1 + \frac{0.04}{365})^{365 t}

Solve for

t, 8900 = 6500 (1 + \frac{0.04}{365})^{365 t} : t = 7.86

Time : 7.86

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