How long will it take $4000 to double if it is invested at 7.85%?
solve for n
8000 = 4000(1.0785)^n
2 = (1.0785)^n
you will have to use logs
To calculate how long it will take $4000 to double at an interest rate of 7.85%, you can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment (in this case, $8000, as you want to double the initial investment)
P = the principal amount (initial investment) ($4000 in this case)
r = annual interest rate (as a decimal) (7.85% = 0.0785 in this case)
n = number of times the interest is compounded per year (assuming it is compounded annually, so n = 1)
t = number of years
Using this formula, you can rearrange it to solve for t:
t = log(A/P) / (n * log(1 + r/n))
Substituting the values into the equation:
t = log(8000/4000) / (1 * log(1 + 0.0785/1))
Now, let's calculate the result:
t = log(2) / (log(1.0785))
Using a calculator, you can find that log(2) ≈ 0.6931 and log(1.0785) ≈ 0.0336:
t ≈ 0.6931 / 0.0336
t ≈ 20.6339
Hence, it will take approximately 20.63 years for $4000 to double at an interest rate of 7.85%.