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

Question

If $5000 is invested at an interest rate k, compounded continuously and grow to $6953.84 in 6 years what is the interest rate and exponential growth function?

Answer 1

6953.84 = 5000 e ^k(6)

6953.84/5000 = e^(6k)

ln (6953.84/5000) = ln e^(6k) = 6k ln e

ln (6953.84/5000) =6k

k = 1/6 ln (6953.84/5000)

k=0.055 (3d.p.)
k = 5.5%

Answer 2

For ok compounding continually the formulation is the exponential e^(nk) the place ok is the each year interest fee and n is the style of years. a) In 6 years, 5000*e^(6k) = 6953.eighty 4 e^(6k) = 6953.eighty 4/5000 = a million.390768 take the organic logarithm of the two factors 6k = 0.329856 ok = 0.054976 or 5.5% b) stability after10 years? using ok = 5.5% or 0.0.5 volume = 5000*e^(10*0.0.5) = volume = $8,666.27 c) How long to double? e^(0.055n) = 2 take Ln of the two factors 0.055n = 0.693147 n = 12.6 years .

Answer 3

5000e^6k = 6953.84
e^6k = 6953.84/5000
e^6k = 1.390768

Take natural log of both sides and solve for k

6k = 0.3298561
k = 0.054976 or about 5.5%

It is interesting to note that calculation of e and calculation of compound interest uses the same formula.

f is the compound interest factor

f = (1 + i)^n where i is interest rate and n is number of corresponding periods, If i is stated as an annual rate and n is not in years, i has to be adjusted accordingly.

To calculate e
(1+ 1/n)^n approaches e (2.71828...)
as n approaches infinity.

Answer 4

N = Ae^kt
6953.84 = 5000e^6k
e^6k = 1.390768
6k = ln 1.90768
k = .054976

rate of interest is 5.4976 %