SOLUTION: Interest Compounded Continuously.
Suppose that Po is invested in a savings account where interest is compounded continuity at 3.1% per year.
A. Express P(t) in terms of Po an
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Question 315557: Interest Compounded Continuously.
Suppose that Po is invested in a savings account where interest is compounded continuity at 3.1% per year.
A. Express P(t) in terms of Po and 0.031.
B. Suppose that $1000 is invested. What is the balance after 1 yr? After 2yrs?
C. When will an investment of $1000 double itself?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
A.
In general:
\ =\ P_oe^{rt})
For a 3.1% interest rate:
\ =\ P_oe^{0.031t})
B. 1 yr:
\ =\ 1000e^{0.031})
B. 2 yr:
\ =\ 1000e^{0.031\,\cdot\,2})
The calculator work is left as an exercise for the student.
C. The investment will double regardless of the value of the initial investment when }{P_o}\ =\ 2)
, which is to say when, in general, 
, or specifically for this problem, 
.
Take the natural log of both sides:
\ =\ \ln(2))
Use \ =\ n\log_b(x))
:
\ =\ \ln(2))
Use \ =\ 1)
)
Divide by 
}{r})
Is the general solution. The specific solution and the resulting arithmetic are left as an exercise for the student.
In a still more general case, the time, 
, to multiply the principal amount by 
, with interest rate, , compounded continuously is:
}{r})
By the way, this was 4 problems in one post. You are allowed only 1 problem per post. Read the instructions, that's why they are there.
John
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