SOLUTION: A person invests 600.00 at 5.64% compounded continously. How long will it take for the money to double. I set it up: {{{1200 = 600e^(.0564x)}}} I am having a hard time solv

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Question 108111: A person invests 600.00 at 5.64% compounded continously. How long will it take for the money to double.
I set it up:
I am having a hard time solving it. Any help would be greatly appreciated.

Found 2 solutions by scott8148, bucky:
Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
dividing by 600 gives 2=e^(.0564x) ... taking natural log gives ln(2)=.0564x ... so x=ln(2)/.0564
Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!
You have set the equation up correctly as:
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Simplify this equation a little by dividing both sides by 600. When you do that you get:
.

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Next take the natural logarithm of both sides to get:
.

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On the right side you can remove the exponent and use it as the multiplier of the natural logarithm.
This is in accordance with one of the rules for logarithmic operations.
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The result is:
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But the natural logarithm of e is 1. Substituting this value for ln(e) results in:
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Use a calculator to find that ln(2) = 0.69314718. Substitute that value for the left side
of the equation to get:
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Solve for x by dividing both sides by 0.0564 to get:
.

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This means that with continuous compounding of interest at an annual interest rate of 0.0564
(that is 5.64%) the amount of money originally invested will double in 12.28984 years.
The fractional part of 0.28984 is equivalent to approximately 3.5 months. So the money doubles
in 12 years and 3.5 months.
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Hope this helps you to understand the problem and how to go about solving it.
.

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