document.write( "Question 1040540: Suppose $7,100 is invested in an account at an annual interest rate of 3.4% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size? Answer: ________ \n" ); document.write( "

Algebra.Com's Answer #655367 by Aldorozos(172) $\"\"$ $\"About$

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F= future value
\n" ); document.write( "P=present value
\n" ); document.write( "F=p*e*rt
\n" ); document.write( " Read below to understand continuous compounding
\n" ); document.write( "http://cs.selu.edu/~rbyrd/math/continuous/
\n" ); document.write( "F=2p\r
\n" ); document.write( "\n" ); document.write( "2p =p * e^.034t
\n" ); document.write( "Divide both sides by p\r
\n" ); document.write( "\n" ); document.write( "2= e^.034t t is the time
\n" ); document.write( "You have to know properties of log and Ln to solve this\r
\n" ); document.write( "\n" ); document.write( "Get Ln of both sides\r
\n" ); document.write( "\n" ); document.write( "Ln of 2 = Ln of e^.034t
\n" ); document.write( "Ln of 2 = .034r* Ln of e
\n" ); document.write( "Using your calculator you will find Ln of 2 is .693 and Ln of e is one
\n" ); document.write( ".693 = .034 t
\n" ); document.write( "T = time = years = .693/.034. = 20.4 which means it takes around 20.4 years for the money to double. \n" ); document.write( "

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