A.How long will it take for 3,700 double if it is invested at 9% annual interest compounded 6 times a year.
B. Compounded continuously, it would only take blank years.
(1+.09/6)^(6t) = 2
e^.09t = 2
To calculate the time it will take for an investment to double, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (double the initial investment)
P = the initial investment
r = the annual interest rate (in decimal form)
n = the number of times the interest is compounded per year
t = the time in years
Let's solve for time (t) in the first scenario:
A = P(1 + r/n)^(nt)
2P = P(1 + 0.09/6)^(6t) (Substituting the given values)
2 = (1.015)^(6t) (Dividing both sides by P)
Now, we can take the logarithm of both sides to solve for t:
log(2) = log((1.015)^(6t))
log(2) = 6t * log(1.015) (Using the logarithmic property)
t = log(2) / (6 * log(1.015)) (Dividing both sides by 6 * log(1.015))
Using a calculator, we can evaluate the right side of the equation:
t ≈ log(2) / (6 * log(1.015)) ≈ 12.2 years
Therefore, it would take approximately 12.2 years for an investment of $3,700 to double at 9% annual interest compounded 6 times a year.
Now let's solve for the time in years (t) when the interest is compounded continuously. In this scenario, we use the formula:
A = P * e^(rt)
Where:
e is a mathematical constant approximately equal to 2.71828
To double the investment, we have:
2P = P * e^(rt)
2 = e^(rt)
Taking the natural logarithm (ln) of both sides:
ln(2) = ln(e^(rt))
ln(2) = rt * ln(e) (Simplifying ln(e) to 1)
rt = ln(2) (Simplifying the natural logarithm)
To solve for time (t), we divide both sides by the annual interest rate (r):
t = ln(2) / r
Using a calculator, we can evaluate the right side of the equation:
t ≈ ln(2) / r ≈ 0.693 / r
Therefore, when the interest is compounded continuously, it would take approximately 0.693 divided by the annual interest rate (r) in years to double the investment.