# A sum of money doubles in 3 years at compound interest, compounded annually. It will become 4 times of itself in

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

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

Where: A = the amount of money after t years P = the initial principal (the sum of money) r = the annual interest rate (as a decimal) n = the number of times the interest is compounded per year t = the time in years

We know that the sum of money doubles in 3 years, so we can set up the following equation:

2P = P(1 + r/1)^(1*3)

Simplifying this equation, we get:

2 = (1 + r)^3

Taking the cube root of both sides, we get:

1 + r = 1.2599

r = 0.2599 or 25.99%

Now we need to find out how long it will take for the sum of money to become 4 times itself. Let’s call this time t. Then we have:

4P = P(1 + r/1)^(1*t)

Simplifying this equation, we get:

4 = (1 + r)^t

Taking the logarithm of both sides (with base 10, for example), we get:

log(4) = t*log(1 + r)

Solving for t, we get:

t = log(4) / log(1 + r)

Plugging in the value of r we found earlier, we get:

t = log(4) / log(1.2599)

Using a calculator, we find:

t ≈ 9.06

Therefore, it will take about 9.06 years for the sum of money to become 4 times itself at compound interest, compounded annually.

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