A person places $38100 in an investment account earning an annual rate of 4%, compounded continuously. Using the formula V, equals, P, e, start superscript, r, t, end superscriptV=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 9 years.
Question
A person places $38100 in an investment account earning an annual rate of 4%, compounded continuously. Using the formula V, equals, P, e, start superscript, r, t, end superscriptV=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 9 years.
Solution
To solve this problem, we will use the formula for continuous compound interest, which is V = Pe^(rt).
Here's the step-by-step solution:
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Identify the variables in the problem:
- P (principal initially invested) = $38100
- r (rate of interest) = 4% = 0.04 (as a decimal)
- t (time in years) = 9 years
- e (base of a natural logarithm) = approximately 2.71828
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Substitute the values into the formula:
- V = 38100 * e^(0.04 * 9)
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Calculate the exponent first (0.04 * 9 = 0.36), then raise e to that power:
- e^0.36 = approximately 1.4338
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Multiply the result by the principal:
- V = 38100 * 1.4338
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Calculate the final value:
- V = approximately $54616.78
So, after 9 years, the account will have about $54616.78.
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