Question

Suppose that an amount in Kina, is invested in a private financial institution, with interest compounded continuously at 8% per year.

a). Write the equation in terms of P₀ and 0.08 where P₀ is the starting amount invested. And the final balance in the account is denoted with variable P

b). Suppose that K2000 is invested. What is the total amount in the account after 3years?

c). How many years will it take to have more then the invested amount.

Answer 1

(a)

As we are finding the final amount, that will be in future, thus we will use the future value formula:

$FV = PV \times (1+r)^n$

Where:

FV = future value

PV = present value

r = interest rate

n = number of years.

using the formula from above, we can write the expression for question as following

$P = P0(1+0.08)^n$

 (b)

To find the value of investment after 3 years, we will use the formula from step-1

$FV = PV \times (1+r)^n$

However, first we will have to find effective annual interest rate, as rate given is continuously compounding.

effective interest rate $= e^r -1$

So,

$e^{0.08 }- 1 = 1.08329 - 1 = 0.08329$

Now we will use the FV formula,

$FV = 2000 (1+0.08329)^3$

$= 2000 (1.271249)$

$= 2542.5$

Thus,

If K2000 is invested, account balance after 3 years will be K2,542.5.

 (c)

To find it, we will use the following expression

$P >= 2P0$

as we determined the value of in step-1,

$P0 (1+0.08)^n = 2P0$

So, we can write it as following (P0 is cancelled out 

(1+0.08)ⁿ = 2

To find it we will use the logarithm (ln = log normal) 

So,

$N ln1.08 = ln2$

finding natural logarithm for both numbers 

$0.07696 = 0.69315\\ = \frac{0.07696 }{ 0.69315}\\ = 9.00647\\ = 9.007$

as the number is not complete, money will be same in 10th year 

Thus, 

It will take more than 9 years (10 years) to have more money than invested in the account.