document.write( "Question 624286: find the time when sum of $2400 increases by 1/5 of every year amounts to $3600.\r
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document.write( "we need to know the rate of interest as given as per my opinion is the rate is 1/5 or what\r
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document.write( "Thank you
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document.write( "Joesph \n" );
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Algebra.Com's Answer #392624 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! find the time when sum of $2400 increases by 1/5 of every year amounts to $3600. \n" ); document.write( "** \n" ); document.write( "1/5 is the same as 20% interest so we can use the compound interest formula: \n" ); document.write( "A=P(1+i)^n, P=initial investment, i=interest rate per period, A=amount after n periods \n" ); document.write( ".. \n" ); document.write( "For given problem: \n" ); document.write( "P=$2400 \n" ); document.write( "i=20% \n" ); document.write( "n=to find \n" ); document.write( "A=$3600 \n" ); document.write( ".. \n" ); document.write( "3600=2400(1+.2)^n \n" ); document.write( "3600/2400=1.2^n \n" ); document.write( "1.5=1.2^n \n" ); document.write( "take log of both sides \n" ); document.write( "log1.5=nlog1.2 \n" ); document.write( "n=log1.5/log1.2 \n" ); document.write( "n≈2.2 yrs \n" ); document.write( " \n" ); document.write( " |