How long it will take for an investment of 2000 dollars to double in value if the interest rate is 9.5 percent per year, compounded continuously?

Continuously Compounded Interest

Continuously compounded interest is interest that is, hypothetically, computed and added to the balance of an account every instant. This is not actually possible, but continuous compounding is well-defined nevertheless as the upper bound of "regular" compound interest. The formula, is sometimes called the shampoo formula (Pert®)

Thus A=Pe ^rt

where e = napier's number or euler's constamt

p = principal/initial investment

r=annual interest rate as a decimal

t = number of years

a = amount of money after t years

thus we plug in to get 4000=2000*e ^9.5t

= 4000 ~= 5436.6^9.5t.

You can solve for t from here

To determine how long it will take for an investment of $2000 to double in value with a continuous compounding interest rate of 9.5 percent per year, we can use the formula:

Doubling time (in years) = ln(2) / (r * t)

Where:
- ln(2) is the natural logarithm of 2 (approximately 0.6931)
- r is the annual interest rate (in decimal form, so 9.5% becomes 0.095)
- t is the time period (in years)

Substituting the given values into the formula:

Doubling time (in years) = ln(2) / (0.095 * t)

We want to find the value of t, the doubling time. Rearranging the formula to solve for t:

t = ln(2) / (0.095 * Doubling time)

Now, we can calculate t using the given values:

t = ln(2) / (0.095 * Doubling time)
t = 0.6931 / (0.095 * Doubling time)

To find the doubling time, we need to divide the natural logarithm of 2 (approximately 0.6931) by the product of 0.095 and the doubling time.

Let's solve for the doubling time.

To determine the time it will take for an investment to double in value, we can use the formula for continuous compounding:

A = P * e^(rt)

Where:
A = final amount (double the initial investment)
P = initial investment
e = mathematical constant approximately equal to 2.71828
r = interest rate (in decimal form)
t = time

In this case, the initial investment (P) is $2000, and the interest rate (r) is 9.5% per year, or 0.095 as a decimal.

We need to solve for t, which represents the time it will take for the investment to double. Rearranging the formula, we have:

2P = P * e^(rt)

Dividing both sides by P, we get:

2 = e^(rt)

To isolate t, we take the natural logarithm of both sides:

ln(2) = ln(e^(rt))

Using the property of logarithms (ln(x^y) = y * ln(x)), we simplify:

ln(2) = rt * ln(e)

Since ln(e) equals 1, the equation further simplifies:

ln(2) = rt

Now, we can solve for t by dividing both sides by r:

t = ln(2) / r

Substituting the values, we find:

t ≈ ln(2) / 0.095

Calculating this, we get:

t ≈ 7.28 years

Therefore, it will take approximately 7.28 years for an investment of $2000 to double in value with a continuously compounded interest rate of 9.5% per year.