How long to the nearest year, will it take an investment to double if it is continuously compounded at 4% per year
Find the amount accumulated in the sinking fund it $450 is deposited monthly for 20 years at 9% per year? (assume end_of_period deposits and compounding at the same intervals as deposits. Round your answer to the nearest cent.)
To find out how long it will take for an investment to double with continuous compounding, you can use the formula for continuous compound interest:
A = P * e^(rt)
Where:
A = the final amount (double the initial investment)
P = the principal amount (initial investment)
e = Euler's number (approximately equal to 2.71828)
r = annual interest rate (in decimal form)
t = time (in years)
In this case, we want to find how long it takes for the investment to double, so A is equal to 2P. Also, the interest rate is 4%, which is equivalent to 0.04 in decimal form. Plugging in these values, the equation becomes:
2P = P * e^(0.04t)
We can cancel out P on both sides:
2 = e^(0.04t)
Now, we can solve for t. Take the natural logarithm (ln) of both sides:
ln(2) = ln(e^(0.04t))
Using the property of logarithms that ln(a^b) = b * ln(a), the equation becomes:
ln(2) = 0.04t * ln(e)
Since ln(e) is equal to 1, the equation simplifies to:
ln(2) = 0.04t
Now, divide both sides by 0.04:
ln(2)/0.04 = t
Using a calculator, we can find that ln(2)/0.04 is approximately 17.328.
Therefore, it will take around 17.328 years (rounded to the nearest year) for the investment to double with continuous compounding at a rate of 4% per year.