$3,600 is invested in an account at an annual interest rate of 4.8% compounded
continuously. How long, to the nearest tenth of a year, will it take the investment to double in
size??
Any help would be appreciated.
The initial amount does not matter.
You want e^0.048t = 2
t = ln2/0.048 = ?
14.4
15 years
nah u suck mine
To find the time it takes for an investment to double in size, we can use the formula for continuous compounding:
A = P * e^(rt)
Where:
A = the final amount (double the initial investment)
P = the initial amount ($3,600)
r = the annual interest rate (4.8% or 0.048)
t = the time in years we want to find
e = Euler's number (approximately 2.71828)
In this case, we know that the final amount is twice the initial investment, so A = 2P. Let's substitute the values into the formula:
2P = P * e^(rt)
Canceling out the P on both sides:
2 = e^(rt)
Taking the natural logarithm of both sides:
ln(2) = ln(e^(rt))
Using the property of logarithms:
ln(2) = rt * ln(e)
The ln(e) term simplifies to 1, so we have:
ln(2) = rt
Now, we can rearrange the equation to solve for t:
t = ln(2) / r
Plugging in the given values:
t = ln(2) / 0.048
Using a calculator, we find:
t ≈ 14.418
Therefore, it will take approximately 14.4 years (rounded to the nearest tenth of a year) for the investment to double in size when compounded continuously.