An investment grows ar a rate of 2.5% each year compounded continuously. Approximately how long will it take the investment to double?
dx/dt = .025 x
dx/x = .025 dt
ln x = .025 t + c
x = e^(.025 t+c) = e^c e^.025 t
so x = C e^.025 t
at t= 0, x = C e^0 = C
when is x = 2C ???
2C = C e^.025 x
e^.025 x = 2
.025 x = ln 2 = .6931
x = 27.7 years
To determine approximately how long it will take for the investment to double, we can use the continuous compound interest formula:
A = P * e^(rt)
Where:
A = Final amount (in this case, double the initial investment)
P = Principal amount (initial investment)
e = Euler's number (approximately 2.71828)
r = Annual interest rate (2.5% or 0.025 as a decimal)
t = Time (in years)
We need to solve for "t" in this case. Rearranging the formula, we have:
2P = P * e^(0.025t)
Dividing both sides by P, we get:
2 = e^(0.025t)
To isolate "t", we take the natural logarithm (ln) of both sides:
ln(2) = 0.025t * ln(e)
Since ln(e) is equal to 1, the equation simplifies to:
ln(2) = 0.025t
Now, we can solve for "t" by dividing both sides by 0.025:
t = ln(2) / 0.025
Using a calculator, we find:
t ≈ 27.725 years
Therefore, it will take approximately 27.725 years for the investment to double at a continuous compounded interest rate of 2.5% per year.