Solve the solution.

Question

Solve the solution.

 

Suppose that $29,000 is deposited in an account and the balance increases to $31,258.64 after 1.5 years. how long will it take for the account to grow to $49,516.00? Assume continuous compounding.

 

 

it will take about ? years for $29,000 to grow to $49,516.00

 

(round to the nearest tenth as needed)

Answer 1

Keith,

         Continuous compounding is modeled by A(t)=A₀ e^kt where A₀ is the amount at t=0, k is the growth rate in yrs^-1 t is the time in years. You are given that A(1.5) =31,258.64 and A₀ = 29000. So we now need to calculate k. So take Ln of each side of the model equation gives Ln(A/A₀) = kt so k = (1/t)Ln(A/A₀) putting in all the numbers gives k= (1/1.5)Ln(31,258.64/29,000) = .05 yr^-1 

So now we have A(t)=29000e^.05t if you put 49516.00=29000e^.05t and solve for t. I got 10.7 years. So in 10.7  years at 5%/yr continuous compounding $29,000 will grow to $49,516.

 

Regards

Jim

Answer 2

A = P e^(rt)

 

$29,000 is deposited in an account and the balance increases to $31,258.64 after 1.5 years.

 

31258.64 = 29000 e^(r(1.5))

 

ln(31.25864/29) = 1.5 r

 

r = ln(31.25864/29) / 1.5

 

How long will it take for the account to grow to $49,516.00?

 

49516 = 29000 e^(t(ln(31.25864/29) / 1.5))

 

ln(49.516/29) = t(ln(31.25864/29) / 1.5)

 

t = 1.5 ln(49.516/29) / ln(31.25864/29)

 

t ≈ 34.14897098909949 ≈ 34.1 years

Answer 3

Assuming Compound Semiannually:

 

    29 000 * ( 1 + r/ 2 ) ^3 = 31, 258 .64

 

     ( 1 + r ) ^3 = 31, 258.64 / 29000

 

      3 log( 1 +r/2) = log 31,258 / 29000

 

        log ( 1 + r/2) = 0.018/3 =0.006

 

                   1 +r/2 = 10 ^( 0.006 0= 1.013)

 

                  r/2 = .043

              r = 0.086

 

          29000 * ( 1 + 0.086/2) ^ n = 49 516

 

         n log ( 1 + 0.043) = log ( 49516/ 29000)

 

         n =  log( 49516/29000) / log 1.043 =

 

         n = 12.7  Semiannual = 6.35 year