document.write( "Question 331898: please help\r
\n" ); document.write( "\n" ); document.write( "Suppose that you invest $3,000 into an account that pays 4.3% interest compounded continuously, find to the nearest tenth of a year when the account will be worth $9,000.\r
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Algebra.Com's Answer #237896 by jrfrunner(365) $\"\"$ $\"About$

You can put this solution on YOUR website!
F=Future value
\n" ); document.write( "r=annual interest rate
\n" ); document.write( "t= number of years
\n" ); document.write( "I=Initial investment
\n" ); document.write( "======
\n" ); document.write( "The formula for continuously compounded value is \r
\n" ); document.write( "\n" ); document.write( " $\"F=I%2Ae%5E%28rt%29\"$
\n" ); document.write( "given F=9000, I=3000, r=4.3%, t=unknown\r
\n" ); document.write( "\n" ); document.write( " $\"9000=3000e%5E%280.043%2At%29\"$ solve for t\r
\n" ); document.write( "\n" ); document.write( "ln(9000/3000)=0.043*t
\n" ); document.write( "ln(3)/0.043=t\r
\n" ); document.write( "\n" ); document.write( "can you finish this?\r
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