If $5000 is invested at an interest rate k, compounded continuousy and grow to $6953.84 in 6 years what is the

Question

If $5000 is invested at an interest rate k, compounded continuously and grow to $6953.84 in 6 years what is the balance after ten years? and whats the doubling time?

I got the interest rate at k = 5.5%

Answer 1

first find k

6953.84 = 5000 *(1+k)^6
(1+k)^6 = 1.390768
(1+k) = 1.0565
k is 5.65% pa

so the equation is
b = 5000 *1.565^t

where t= time in years
and b= balance at the time

balance after 10 years

b=5000 *1.0565^10
b=8664.19
(8664 to 4 significant figures)

doubling time

k^t=2
1.0565^t=2
t= log base 1.0565 of 2
t= ln(2)/ln(1.0565)
t= 12.6115 years


check

5000* 1.0565^12.611 = 10000

(12.61 years to 4 significant figures)

remember you only need to give significant figures in your answer, your working should use numbers that are accurate enough to give the correct answer

Answer 2

For k compounding continuously the formula is

the exponential e^(nk)
where k is the annual interest rate and n is the number of years.

a) In 6 years,
5000*e^(6k) = 6953.84
e^(6k) = 6953.84/5000 = 1.390768

take the natural logarithm of both sides
6k = 0.329856
k = 0.054976 or 5.5%

b) Balance after10 years?
Using k = 5.5% or 0.055
amount = 5000*e^(10*0.055) =
amount = $8,666.27

c) How long to double?
e^(0.055n) = 2
take Ln of both sides
0.055n = 0.693147

n = 12.6 years
.

Answer 3

A = P(1 + k)^10

A = 5000( 1 + .055)^10

A = 5000(1.055)^10 ==> amount after 10 years

Doubling time???
72/5.5 = 13 years (approximately)

or

t = ln2/r
t = .693/.055
t = 12.60 or 13 years

Answer 4

k = 5.6515%
in 10 yrs = $8,664.19

it will double in yr 13