SOLUTION: If interest is compounded continuously, at what annual rate will a principal double in 20 years? Give the answer as a percentage correct to two decimal places.

Question 556132: If interest is compounded continuously, at what annual rate will a principal double in 20 years? Give the answer as a percentage correct to two decimal places.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
The formula for continuously compounded interest is:
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In which P is the future value of the investment, C is the initial deposit, e is the base of the natural logarithms, r is the annual interest rate expressed as a decimal, and t is the number of years that the money is to be invested.
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In this case, the initial investment is C and the future value is to be twice C. So we can substitute 2 time C for P and the equation becomes:
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If we then divide both sides of the equation by C, the two C factors disappear and the equation becomes:
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Since the investment is to be for 20 years, we can substitute 20 for t and the equation is then:
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Then we take the natural logarithm (base e) of both sides and we have:
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But the rules of logarithms say that an exponent can be brought out as a multiplier of the logarithm. So on the right side we can bring the 20*r out as a multiplier of the logarithm and the equation becomes:
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But the natural logarithm of e is equal to 1. So we can replace ln(e) by 1 and the equation reduces to:
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Solve for r by dividing both sides by 20 to get:
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On a calculator we can find that ln(2) equals 0.69314718. When we substitute that value for ln(2) the equation becomes:
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Doing this division on a calculator we find that:
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So this is the rate in decimal form that will double a continuously compounded investment in 20 years. You can convert this to percent by moving the decimal two places to the right to get 3.4657359%. Rounded to two decimal places the percentage interest rate would be 3.47%.
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Hope this helps to clarify how you could do this problem and provides some insight into working with logarithms to solve exponential problems.
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