You put $8,000 into a savings account that has an annual interest rate of 3%, compounded annually. How many years will it take for your savings to increase to $12,000? Use the formula A = P(1 + (r)/(n))^nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years. Round your answer to the nearest hundredth.

Question

You put $8,000 into a savings account that has an annual interest rate of 3%, compounded annually. How many years will it take for your savings to increase to $12,000? Use the formula  A = P(1 + (r)/(n))^nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.  Round your answer to the nearest hundredth.

Bytelearn · Accepted Answer

Identify values: Identify the values of P, r, n, and A. P = 8000 r = 0.03 n = 1 A = 12000Use formula and solve: Use the formula A = P(1 + (r)/(n))^nt and solve for t. 12000 = 8000(1 + 0.03/1)^(1*t) 12000 = 8000(1.03)^tDivide to isolate: Divide both sides by 8000 to isolate (1.03)^t. 12000 / 8000 = (1.03)^t 1.5 = (1.03)^tTake natural logarithm: Take the natural logarithm (ln) of both sides to solve for t. ln(1.5) = ln((1.03)^t) ln(1.5) = t * ln(1.03)Divide to isolate: Divide both sides by ln(1.03) to isolate t. t = ln(1.5) / ln(1.03) t ≈ 0.405465 / 0.029559 t ≈ 13.72

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