How long will it take $3000 to double if it is invested at 10% annual interest compounded 12 times a year.
A=P(1+r/n)^n*t
6000=3000 (1+.10/12)^12t
6000-3000=3000-3000(1+008333333)^12t
3=1.008333333^12t
In 3=In 1.008333333^12t
In 3/12In 1.008333333=12tIn 1.008333333/12In1.008333333
= In(3)/12In(1.008333333
=1.098/0.099
= 11.0 or 11 years.
2 = [1 + (.1 / 12)]^(12 t)
ln(2) = 12t ln(12.1 / 12)
[ln(2)] / [ln(12.1 / 12)] = 12 t
major error from 6000=3000 (1+.10/12)^12t
to
6000-3000=3000-3000(1+008333333)^12t
which in turn is NOT equal to the next line
3=1.008333333^12t
propter steps:
6000 = 3000(1.0083333)^t, where t is in months
2 =1.0083333^t
take ln of both sides
ln2 = t ln1.083333.
t = ln2/ln1.0083333... = 83.523 months
= appr 6.96 years ----> which is what Scott's answer is also
To get the answer to this question, we can use the formula for compound interest:
A = P (1 + r/n)^(n*t)
Where:
A = the final amount
P = the initial amount (in this case, $3000)
r = the annual interest rate (in decimal form, 10% = 0.10)
n = the number of times interest is compounded per year (in this case, 12 times)
t = the number of years
We want to find the value of t (number of years) when the initial amount doubles, so the final amount A will be $6000.
Therefore, the equation becomes:
6000 = 3000 (1 + 0.10/12)^(12*t)
Simplifying further, we have:
6000/3000 = (1 + 0.10/12)^(12*t)
2 = (1 + 0.008333333)^12*t
Take the logarithm (ln) of both sides:
ln(2) = ln((1 + 0.008333333)^12*t)
Using the property of logarithms, we can rewrite the equation as:
ln(2) = 12*t * ln(1.008333333)
Finally, solve for t by dividing both sides by 12*ln(1.008333333):
t = ln(2) / (12*ln(1.008333333))
Evaluating this expression on a calculator, we get approximately t = 11.0.
Therefore, it would take approximately 11 years for $3000 to double if it is invested at a 10% annual interest rate compounded 12 times a year.