how long does it take an investment of $8000 to double if it is invested at 7% interest compounded continuously. round to nearest year.
e^(1.07t) = 2
Future value = Principle × e^(interest rate × time) =
16,000 = 8,000e^(0.07t)
2 = e^(0.07t)
ln (2) = 0.07t
0.693 = 0.07t
t = 0.693/0.07 = 9.90 years which rounds to 10 years.
QED
Sorry about my typo. e^.07t is correct
To solve this problem, we can use the formula for continuous compound interest:
A = P * e^(rt),
where:
A is the final amount (in this case, double the initial investment)
P is the initial investment ($8000)
e is the mathematical constant approximately equal to 2.71828
r is the annual interest rate (7% or 0.07, in decimal form)
t is the time in years
Let's plug in the values and solve for t:
2P = P * e^(0.07t)
To simplify the equation and cancel out the P on both sides:
2 = e^(0.07t)
Next, take the natural logarithm (ln) of both sides to isolate the exponent:
ln(2) = ln(e^(0.07t))
Using the natural logarithm property (ln(e^x) = x), we get:
ln(2) = 0.07t
Now, divide both sides by 0.07 to solve for t:
t = ln(2) / 0.07
Calculating this value:
t ≈ 9.90 years
Therefore, it takes approximately 9.90 years for an investment of $8000 to double at an annual interest rate of 7% compounded continuously. Rounded to the nearest year, it would take 10 years.