Question

You invest $2500 in an account at interest rate r, compounded continuously. Find the time required for the amount to double and triple. (Round your answers to twodecimal places.)r = 0.0465(a) doubleyr(b) tripleуг

Answer 1

To determine the time required for the investment to double and triple with continuous compounding at a rate of 0.0465, we use the formula A = Pe^(rt). It takes approximately 14.91 years for the investment to double and approximately 22.39 years to triple.

Step-by-step explanation:

When an investment is compounded continuously, the formula used is A = Pert, where A is the amount of money accumulated after t years, including interest. P is the principal amount (the initial amount of money), r is the annual interest rate (decimal), and e is the base of the natural logarithm, approximately equal to 2.71828. In this case, the principal is $2500 and the interest rate r is 0.0465.

To find the time t required for the investment to double, we set A to $5000, which is twice the original investment. To solve for t, we use the formula:

5000 = 2500e(0.0465t)

Dividing both sides by 2500 gives us:

2 = e(0.0465t)

We then take the natural logarithm of both sides:

ln(2) = 0.0465t

And solve for t:

t = ln(2)/0.0465

Using a calculator, we find that t ≈ 14.91 years.

Similarly, to find the time t for the money to triple, we set A to $7500 (three times the original investment) and solve:

7500 = 2500e(0.0465t)

Dividing both sides by 2500 gives us:

3 = e(0.0465t)

Taking the natural logarithm of both sides:

ln(3) = 0.0465t

And solving for t:

t = ln(3)/0.0465

Calculating t gives us approximately 22.39 years.

Answer 2

P= 2500

r= 0.0465

A= P. e^(rt)

Where:

A = final mount

P = principal investment

r= rate

T= time (years)

To double it:

2 x 2500 = 2500 e^(0.0465 x t)

(2 x2500)/2500= e^(0.0465 x t)

2= e^(0.0465 x t)

Ln 2 = Ln e^(0.0465 x t)

Ln2 =0.0465t (Ln e)

Ln2 = 0.0465 t

t = Ln2 /0.0465

t= 14.91 years

Triple:

3 x2500 = 2500 e^(0.0465 t)

(3 x 2500 ) /2500 = e^(0.0465 t)

3 = e^(0.0465 t)

ln3 = ln e^(0.0465 t)

ln3 = 0.0465t Lne

ln3= 0.0465t

ln3/0.0465 =t

t= 23.63 y