document.write( "Question 134074: If $10000 is invested at 6% compounded continously, in approximately how many years will the amount have grown to $15000? \n" ); document.write( "
Algebra.Com's Answer #98115 by nycsharkman(136)![]() ![]() ![]() You can put this solution on YOUR website! We use the formula A = Pe^(rt)where \r \n" ); document.write( "\n" ); document.write( "A = final balance \r \n" ); document.write( "\n" ); document.write( "P = original investment \r \n" ); document.write( "\n" ); document.write( "r = the annual interest rate (as a decimal) \r \n" ); document.write( "\n" ); document.write( "t = investment time (in years) \r \n" ); document.write( "\n" ); document.write( "e = a special \"calculator number\" that serves as the base of the \n" ); document.write( "nalural logarithm.\r \n" ); document.write( "\n" ); document.write( "Your question:\r \n" ); document.write( "\n" ); document.write( "If $10000 is invested at 6% compounded continously, in approximately how many years will the amount have grown to $15000?\r \n" ); document.write( "\n" ); document.write( "Let A = 15000\r \n" ); document.write( "\n" ); document.write( "15000 = 10000e^(0.06t)\r \n" ); document.write( "\n" ); document.write( "Divide both sides by 10000.\r \n" ); document.write( "\n" ); document.write( "1.5 = e^(0.06t)\r \n" ); document.write( "\n" ); document.write( "Rewrite as a log.\r \n" ); document.write( "\n" ); document.write( "0.06t = In(1.5)\r \n" ); document.write( "\n" ); document.write( "Solve for t.\r \n" ); document.write( "\n" ); document.write( "t = In(1.5)/0.06\r \n" ); document.write( "\n" ); document.write( "t = 6.757751802\r \n" ); document.write( "\n" ); document.write( "Round 6.757751802 to the nearest ones place and we get\r \n" ); document.write( "\n" ); document.write( "t is about 7 years.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |