An investment grows ar a rate of 2.5% each year compounded continuously. Approximately how long will it take the investment to double?

User Icon for Damon Damon answered
13 years ago

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

User Icon for Explain Bot Explain Bot answered
11 months ago

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.