document.write( "Question 1209277: A man makes a deposit of $50,000 into an account that pays an annual interest of 5%. How much money will be in the account after 10 years if:
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document.write( "I. The interest is compounded continuously
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document.write( "II. How long will it take for the money to triple itself if interest is compounded continuously? \n" );
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Algebra.Com's Answer #848006 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Part I\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "P = 50000 = deposit amount \n" ); document.write( "e = special constant 2.718 approximately \n" ); document.write( "r = 0.05 = decimal form of the interest rate \n" ); document.write( "t = 10 years\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A = P*e^(r*t) \n" ); document.write( "A = 50000*e^(0.05*10) \n" ); document.write( "A = 82436.06353501 \n" ); document.write( "A = 82436.06\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Part II\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The man deposits P dollars and wants it to triple to 3P dollars. \n" ); document.write( "You could use P = 50,000 from earlier, but this also works for any positive real number.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The r value is the same as before. \n" ); document.write( "The goal is to solve for variable t.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A = P*e^(r*t) \n" ); document.write( "3P = P*e^(0.05*t) \n" ); document.write( "3 = e^(0.05*t) \n" ); document.write( "Ln(3) = Ln( e^(0.05*t) ) \n" ); document.write( "Ln(3) = 0.05*t*Ln( e ) \n" ); document.write( "Ln(3) = 0.05*t*1 \n" ); document.write( "Ln(3) = 0.05*t \n" ); document.write( "t = Ln(3)/0.05 \n" ); document.write( "t = 21.972246\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answers: \n" ); document.write( "I. $82,436.06 \n" ); document.write( "II. 21.972246 years approximately \n" ); document.write( " \n" ); document.write( " |