# Thread: Continuous Compound Interest problem

1. ## Continuous Compound Interest problem

So I know this is the formula,
A
= Pe rt

To what final amount will an initial investment of $5,000 grow if the interest rate is 4% per year compounded continuously for 21 years? Write your answer rounded to the nearest cent. im getting a number of 7.994 but it doesnt seem right. 2. Originally Posted by av8or91 So I know this is the formula, A = Pe rt To what final amount will an initial investment of$5,000 grow if the interest rate is 4% per year compounded continuously for 21 years? Write your answer rounded to the nearest cent.

im getting a number of 7.994 but it doesnt seem right.
Of course it's not right as \$5000 > 7.994 and the problem implies that the amount increases

$P = 5000e^{21 \times 0.04} = \11,581.83$

3. I dont understand what im doing wrong. 5000e^(.04)(21) Then where do I proceed. P=.84ln5000e?

4. Originally Posted by av8or91
I dont understand what im doing wrong.
You're doing nothing wrong,

evaluate this!

Originally Posted by av8or91
5000e^(.04)(21)

5. Maybe thats my issue, im just evaluating it wrong. .84ln5000e does the equal .84ln13590? Then what I am doing is taking the natural log of 13590 mulitply by .84 but no thats not right.

6. Originally Posted by av8or91
Maybe thats my issue, im just evaluating it wrong. .84ln5000e does the equal .84ln13590? Then what I am doing is taking the natural log of 13590 mulitply by .84 but no thats not right.
You don't need to use the natural log at all.

You're trying to find A given P, r and t. You can directly substitute the given values into the equation you gave in the OP to find a value for A

7. Maybe im a little stupid but thats why im here for help, . Can you please show me how you are not using any natural logs and how you did it. Much appreciated.

8. There are no logs in your original equation therefore consider

$A = Pe^{rt}$

inputing the information given

$A= 5000e^{0.04\times 21}$

$A = 5000e^{0.84}$

$A = 5000\times 2.316$

$A = 11,581.83$

that is all.

9. Ok Thanks. I see that now but when would you use ln

10. Originally Posted by av8or91
Ok Thanks. I see that now but when would you use ln
In the problem you provided!?!

11. In this problem you have all variables given (P, r and t)

In this example you would use ln:

How long will it take to double your investment if the interest would be 6% compounded continuousely?

A= P * e^rt

P = 1000
A = 2000
r = 0.06

2000 = 1000 * e^0.06*t
2 = e^0.06*t
ln2 = ln e^0.06*t --> ln e^x = x
ln2 = 0.06*t
t= ln2/0.06