A sum of $5000 is be invested in a bank. if the annual interest is 10% and compounded monthly, how long will it take for the original investment to double.

i = .10/12 = .008333..

n = ? months

5000(1.008333..)^n = 10000
1.008333..^n = 2
n log 1.0083333..= log2
n = appr 83.5 months
or 6 years and appr 11.5 months

To determine how long it will take for the original investment to double, we can use the formula for compound interest.

The formula for compound interest is:

A = P(1 + r/n)^(nt)

Where:
A = the final amount (in this case, double the original investment)
P = the principal amount (the original investment)
r = the annual interest rate (as a decimal)
n = the number of times that interest is compounded per year
t = the number of years

In this case, the original investment is $5000, the annual interest rate is 10% (0.10 as a decimal), and interest is compounded monthly. So, we have:

A = $5000 * (1 + 0.10/12)^(12t)

We want to find the value of t when A is equal to 2P (double the original investment):

2P = $5000 * (1 + 0.10/12)^(12t)

Simplifying the equation, we have:

2 = (1 + 0.10/12)^(12t)

Now, by taking the natural logarithm (ln) of both sides of the equation, we can isolate t:

ln(2) = 12t * ln(1 + 0.10/12)

Finally, divide both sides of the equation by 12 ln(1 + 0.10/12) and solve for t:

t = ln(2) / (12 * ln(1 + 0.10/12))

Using a calculator, we can find the value of t, which is approximately 7.27 years.

Therefore, it will take approximately 7.27 years for the original investment to double when the annual interest is 10% and compounded monthly.