SOLUTION: How much more money is earned in 10 years if $5,000 is invested at 6% compounded continuously rather than compounded annually? A=P(1+r/m)^n A= 5,000(1+0.06/365)^3650 A- $9110

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Question 541458: How much more money is earned in 10 years if $5,000 is invested at 6% compounded continuously rather than compounded annually?
A=P(1+r/m)^n
A= 5,000(1+0.06/365)^3650
A- $9110.14
A=Pe^rt
A= 5,000e^(0.06)^(10)
A= $9,110.59
This is what I did but even the difference doesn't match any of the answer possibilities.
The possibilities are:
$312.72
$156.36
$78.18
$6,110.59
Answer by jpg7n16(66) About Me  (Show Source):
You can put this solution on YOUR website!
Ah. Your error is in section 1. Note carefully from the problem what two forms of compounding you are comparing: ANNUALLY vs continually. In your step 1, you did the calculation for DAILY compounding. Try this instead:
A=P%2A%28%281%2Br%2Fm%29%5En%29
A=P%2A%28%281%2B.06%2F1%29%5E10%29
A=P%2A%28%281.06%29%5E10%29
You'll find when you rework the problem with this slight change, one of the answers will work for you.
.
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Your part 2 is correct, although one thing in your work is written incorrectly. You have listed:
A= 5,000e^(0.06)^(10)
Which is confusing. It's hard to tell which piece is being raised to the 10th power. If you're required to show your work, to avoid confusion (and any potential docking points) it should be written as any of the following:
A=5000%2A%28%28e%5E.06%29%5E10%29=5000e%5E%28%28.06%29%2A%2810%29%29=5000e%5E%28.60%29
I'd personally go with the last.
A=5000*e^(.60)