If $2000 is invested at 10% pa compounded annually, how many years will it take to double in value?

To solve this problem, we can use the formula for compound interest, which is:

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

Where: A = the amount of money accumulated after n years, including interest. P = principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = time the money is invested for in years

In this case, we want to find out when the initial amount (P) will double. This means that A = 2P. The interest rate (r) is 10% or 0.10, and the interest is compounded annually, so n = 1.

So, we can set up the equation as follows:

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

We can simplify this to:

2 = (1 + 0.10)^t

Now, we need to solve for t. This is an exponential equation, so we'll need to use logarithms to solve it. We can take the natural log (ln) of both sides:

ln(2) = t * ln(1.10)

Then, solve for t:

t = ln(2) / ln(1.10)

Using a calculator, we find that ln(2) is approximately 0.6931 and ln(1.10) is approximately 0.0953.

So, t = 0.6931 / 0.0953 ≈ 7.27 years.

Therefore, it will take approximately 7.27 years for the investment to double if it is invested at an annual interest rate of 10% compounded annually.