How long will it take for $1,200 to double if it is invested at 12% annual interest compounded 6 times a year? Enter in exact calculations or round to 3 decimal places.
It will take years to double.
How long will it take if the interest is compounded continuously?
Compounded continuously, it would only take years
Surely after all the similar questions you have posted, you have some of your own input by now?
To calculate the time it takes for an amount to double with compounding interest, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the future value (in this case, double the initial amount)
P = the principal amount (initial investment)
r = the annual interest rate (as a decimal)
n = the number of times interest is compounded per year
t = the time in years
In this case, we have P = $1,200, r = 12% = 0.12, n = 6, and we want to find t.
Let's rewrite the formula to solve for t:
2P = P(1 + r/n)^(nt)
Dividing both sides by P:
2 = (1 + r/n)^(nt)
Taking the natural logarithm (ln) of both sides:
ln(2) = nt ln(1 + r/n)
n is given as 6 times a year, which means it compounds semi-annually. So we substitute n = 2:
ln(2) = 2t ln(1 + r/2)
Now, let's solve for t:
t = ln(2) / (2 ln(1 + r/2))
Using the given interest rate of 12% = 0.12:
t = ln(2) / (2 ln(1 + 0.12/2))
Calculating this expression, we can find the exact value or round it to 3 decimal places to get an approximation.
For continuously compounded interest, we can use the formula:
A = P * e^(rt)
Where e is Euler's number (approximately equal to 2.71828). In this case, we want to find t when A = 2P:
2P = P * e^(rt)
Dividing both sides by P:
2 = e^(rt)
Taking the natural logarithm (ln) of both sides:
ln(2) = rt ln(e)
Since ln(e) = 1:
ln(2) = rt
Solving for t:
t = ln(2) / r
Using the given interest rate of 12% = 0.12:
t = ln(2) / 0.12
Again, we can calculate the exact value or round it to 3 decimal places for an approximation.