Question

David invested his profit of $50,000 from the sale of his business into an aggressive stock fund. After 15 years, the account had a total of $132,558.36 in it. A. What annual interest rate, compounded continuously, produced this return? (Enter the value of r rounded to three decimal places.). B. E termine the approximate number of years (from the original investment) it will take for the account to grow to $300,000. Round answer to the nearest tenth

Answer 1

Step-by-step explanation:

A. To find the annual interest rate, compounded continuously, we can use the formula:

A = Pe^(rt)

where A is the final amount, P is the principal amount (initial investment), e is the mathematical constant approximately equal to 2.71828, r is the annual interest rate, and t is the time in years.

We are given that P = $50,000, A = $132,558.36, and t = 15 years. We need to solve for r.

Substituting the given values into the formula, we get:

$132,558.36 = $50,000 e^(15r)

Dividing both sides by $50,000 and taking the natural logarithm of both sides, we get:

ln(2.6511672) = 15r

Solving for r, we get:

r = ln(2.6511672) / 15

r ≈ 0.045 (rounded to three decimal places)

Therefore, the annual interest rate, compounded continuously, that produced this return is approximately 0.045 or 4.5%.

B. To determine the approximate number of years it will take for the account to grow to $300,000, we can again use the formula:

A = Pe^(rt)

where A is the final amount we want to reach, P is the initial investment ($50,000), e is the mathematical constant approximately equal to 2.71828, r is the annual interest rate we just found (0.045), and t is the time in years we want to solve for.

Substituting the given values into the formula, we get:

$300,000 = $50,000 e^(0.045t)

Dividing both sides by $50,000 and taking the natural logarithm of both sides, we get:

ln(6) = 0.045t

Solving for t, we get:

t = ln(6) / 0.045

t ≈ 27.6 years (rounded to the nearest tenth)

Therefore, it will take approximately 27.6 years from the original investment for the account to grow to $300,000.

Answer 2

The annual interest rate, compounded continuously, that produced this return is approximately 0.063. It will take approximately 20.2 years for the account to grow to $300,000 from the original investment.

Step-by-step explanation:

To determine the annual interest rate, compounded continuously, we can use the formula:

A = P * e^(rt)

Where:

• A is the final account balance ($132,558.36)
• P is the initial investment ($50,000)
• e is Euler's number (approximately 2.71828)
• r is the annual interest rate
• t is the number of years (15)

Plugging in the given values, we have:

132,558.36 = 50,000 * e^(15r)

Rearranging the equation to solve for r:

e^(15r) = 132,558.36 / 50,000

Using natural logarithms, we can find the value of r:

ln(e^(15r)) = ln(132,558.36 / 50,000)

15r * ln(e) = ln(132,558.36 / 50,000)

15r = ln(132,558.36 / 50,000)

r = ln(132,558.36 / 50,000) / 15

Calculating this value using a calculator, we get:

r ≈ 0.063 (rounded to three decimal places)

Therefore, the annual interest rate, compounded continuously, that produced this return is approximately 0.063.

To determine the approximate number of years it will take for the account to grow to $300,000, we can use the formula:

A = P * e^(rt)

Where:

• A is the future account balance ($300,000)
• P is the initial investment ($50,000)
• e is Euler's number (approximately 2.71828)
• r is the annual interest rate (0.063)
• t is the number of years we want to find

Plugging in the given values, we have:

300,000 = 50,000 * e(0.063t)

Rearranging the equation to solve for t:

e(0.063t) = 300,000 / 50,000

Using natural logarithms, we can find the value of t:

ln(e(0.063t)) = ln(300,000 / 50,000)

0.063t * ln(e) = ln(300,000 / 50,000)

0.063t = ln(300,000 / 50,000)

t = ln(300,000 / 50,000) / 0.063

Calculating this value using a calculator, we get:

t ≈ 20.2 (rounded to the nearest tenth)

Therefore, it will take approximately 20.2 years for the account to grow to $300,000 from the original investment.