Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 169770: Problem:
Suppose $8,000 is invested at interest rate k, compounded continuously, and grows to $11,466.64 in 6 years.

a) Find the interest rate
b) Find the exponential growth function
c) Find the balance after 10 years
d) Find the doubling time.
Thank you.
: Problem:
Suppose $8,000 is invested at interest rate k, compounded continuously, and grows to $11,466.64 in 6 years.

a) Find the interest rate
b) Find the exponential growth function
c) Find the balance after 10 years
d) Find the doubling time.
Thank you.

Answer by Edwin McCravy(2190) About Me  (Show Source):
You can put this solution on YOUR website!
Problem:
Suppose $8,000 is invested at interest rate k, compounded continuously, and grows to $11,466.64 in 6 years.
a) Find the interest rate

A=Pe^(rt) where A=11466.64, P=8000, t=6

11466.64=(8000)e^(r(6))

11466.64=8000e^(6r)
Divide both sides by 8000
1.43333=e^(6r)

Take natural logs of both sides:

ln(1.43333)=ln(e^(6r))
Use fact that log(B,B^A)=A

ln(1.43333)=6r
Divide both sides by 6:
ln(1.43333)/6=r
Use calculator:
.0600000681=r

We'll round that to .06 or matrix(1,2,6,'%')

---------------------

b) Find the exponential growth function

A=Pe^(rt)

Substitute 8000 for P and .06 for r:

A=8000e^(.06t)

Your teacher may want you to write A in
functional notation as A(t)

A(t)=8000e^(.06t).  Ask your teacher.

-------------------

c) Find the balance after 10 years

Substitute 10 for t in

A=8000e^(.06t).  
A=8000e^(.06(10))
A=8000e^(.6)
A=14576.9504  
matrix(1,4,    A, '=', '$', 14576.95 )    

--------

d) Find the doubling time. 

For $8000 to double, it must become $16000,

So we substitute $16000 for A and solve for t

A=8000e^(.06t)  

16000=8000e^(.06t)

Divide both sides by 8000

2=e^(.06t)

Take natural logs of both sides:

ln(2)=ln(e^(.06t))

Use fact that log(B,B^A)=A

ln(2)=.06t

Divide both sides by .06:

ln(2)/6=t

11.55245301

or a little over 11&1/2 years.

Edwin