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.
(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\\\\\n\n= \\frac{0.07696 }{ 0.69315}\\\\\n\n= 9.00647\\\\\n\n= 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.
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