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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\n" ); 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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\n" ); document.write( "Joesph
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Algebra.Com's Answer #392624 by lwsshak3(11628)\"\" \"About 
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find the time when sum of $2400 increases by 1/5 of every year amounts to $3600.
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\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
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\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
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\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
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