Question

Use the model or , where is the future value of dollars invested at interest rate compounded continuously or times per year for years. A investment grows to at interest compounded quarterly. For how long was the money invested

Answer 1

The model for continuous compounding is where is the principal amount, is the annual interest rate, and is the time in years.

The model for quarterly compounding is , where is the principal amount, is the annual interest rate, is the number of times per year the interest is compounded, and is the time in years.

If an investment grows to $1000 at 6% interest compounded quarterly, we can set up the following equation:

$1000 = P(1 + r/n)^(nt)

where is the principal amount, is the annual interest rate, is the number of times per year the interest is compounded, and is the time in years.

Plugging in the given values, we get:

$1000 = P(1 + 0.06/4)^(4t)

Simplifying, we get:

$1000 = P(1.015)^4t

Dividing both sides by P(1.015)^4t, we get:

$1000/P(1.015)^4t = 1

Taking the natural logarithm of both sides, we get:

ln($1000/P(1.015)^4t) = ln(1)

Simplifying, we get:

ln($1000/P(1.015)^4t) = 0

Using the logarithmic property that ln(a/b) = ln(a) - ln(b), we get:

ln($1000) - ln(P(1.015)^4t) = 0

Simplifying, we get:

ln($1000) = ln(P(1.015)^4t)

Using the logarithmic property that ln(a^b) = b*ln(a), we get:

ln($1000) = ln(P) + 4t*ln(1.015)

Simplifying, we get:

ln($1000/P) = 4t*ln(1.015)

Dividing both sides by 4ln(1.015), we get:

t = ln($1000/P)/(4ln(1.015))

Therefore, the money was invested for t years, where t is calculated using the above formula.