Answer:
- 7.30%; $1452.02
- 4.45%; 15.56 years
Explanation:
You want the interest rate and account balance after 10 years if the investment of $700 doubles in 9 1/2 years with interest compounded continuously. And, you want the interest rate and doubling time if a $900 investment has a value of $1405 after 10 years with continuous compounding.
Compound interest
These problems are solved by making use of the compound interest formula for continuously compounded interest. In each case, you fill in the values you know, and solve for the one you don't know.
A = Pe^(rt)
$700
The investment doubles after 9.5 years, so we have ...
2·700 = 700e^(9.5r)
Dividing by 700 and taking natural logs, we find the rate to be ...
ln(2) = 9.5r
r = ln(2)/9.5 ≈ 0.072963 ≈ 7.30%
Using the exact value in the computation of the balance, we get ...
A = 700e^(10·(ln(2)/9.5)) = 700·2^(10/9.5) = 700·2^(20/19)
A = 1452.02
The annual rate is 7.30%; the balance in 10 years is $1452.02.
$900
The balance after 10 years is ...
1405 = 900e^(r·10)
ln(1405/900) = 10r
r = ln(1405/900)/10 ≈ 0.0445398 ≈ 4.45%
As we found above, the relation between r and doubling time is ...
r = ln(2)/t ⇒ t = ln(2)/r
t = ln(2)/0.0445398 ≈ 15.56 . . . . years
The annual rate is about 4.45%; the doubling time is about 15.56 years.
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Additional comment
In the first problem, we could find the account balance after 10 years using the found interest rate. In that case, it would be $1452.56. Instead, we chose the doubling time to be the exact value in the problem.
There is a "rule of 70" or sometimes the "rule of 72" that tells you the approximate doubling time in years (y) for an interest rate of r%. It is ...
y = 70/r
The amount of error depends on the interest rate and the way interest is compounded.
When the interest is compounded continuously, a slightly different rule gives the exact doubling time:
y = 100·ln(2)/r ≈ 69.31472/r
This is effectively the rule we used above.