$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.