1. How long will it take for $900 to double if it is invested at 9% annual interest compounded 3 times a year? Enter in calculations or round to 3 decimal places.
1. It is take________ years to double.
2. How long will if the interest is compounded continuously?
2. Compounded continuously, it would only take ____________ years.
see previous post for the method.
and recall that continuous interest at rate r is
e^rt = 2
To find the answers to these questions, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount
r = the annual interest rate (expressed as a decimal)
n = the number of times interest is compounded per year
t = the number of years
1. To find how long it will take for $900 to double at 9% annual interest compounded 3 times a year, we need to solve for t when A is twice the initial investment:
A = 2P
2P = P(1 + 0.09/3)^(3t)
Now we can cancel out P from both sides of the equation:
2 = (1 + 0.09/3)^(3t)
To solve for t, we can take the natural logarithm of both sides:
ln(2) = ln((1 + 0.09/3)^(3t))
Divide by 3 and rearrange the equation:
(3t) * ln(1 + 0.09/3) = ln(2)
t = ln(2) / (3 * ln(1 + 0.09/3))
Using a calculator, we can calculate the value of t, which is approximately 5.743 years.
So, it will take approximately 5.743 years for $900 to double at 9% annual interest compounded 3 times a year.
2. To find how long it will take for the interest to double if it is compounded continuously, we use the continuous compound interest formula:
A = Pe^(rt)
Where e is Euler's number (approximately 2.71828).
We can rearrange the formula to solve for t:
2P = Pe^(rt)
2 = e^(rt)
Take the natural logarithm of both sides:
ln(2) = rt
t = ln(2) / r
Using the same interest rate of 9% (0.09 as a decimal), we can calculate t:
t = ln(2) / 0.09
Using a calculator, we find the value of t is approximately 7.722 years.
So, if the interest is compounded continuously, it would take approximately 7.722 years for the investment to double.