SOLUTION: Please help with this question...I cant figure it out...thanks! Suppose $5000 is in invested at interest rate k, compounded continuously, and grows to 6954.84 in 6 yr. a)Find

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Question 460256: Please help with this question...I cant figure it out...thanks!
Suppose $5000 is in invested at interest rate k, compounded continuously, and grows to 6954.84 in 6 yr.
a)Find the interest rate.
b)Find the exponential growth function.
c)Find the balance after 10 yr.
d)Find the doubling time.
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
Suppose $5000 is in invested at interest rate k, compounded continuously, and grows to 6954.84 in 6 yr.
a)Find the interest rate.
Think: A = Pe^(kt)
plug in given values to find r:
6954.84 = 5000e^(6k)
1.390968 = e^(6k)
ln(1.390968) = 6k
ln(1.390968)/6 = k
0.055 = k
or
5.5% = k
.
b)Find the exponential growth function.
A = Pe^(kt)
A = 5000e^(0.055t)
.
c)Find the balance after 10 yr.
Set t to 10 and solve for A:
A = 5000e^(0.055t)
A = 5000e^(0.055*10)
A = 5000e^(0.55)
A = $8666.27
.
d)Find the doubling time.
A = 5000e^(0.055t)
Set A to 10000 and solve for t:
10000 = 5000e^(0.055t)
2 = e^(0.055t)
ln(2) = 0.055t
ln(2)/0.055 = t
12.6 years = t