document.write( "Question 252185: A GIC pays 6% per annum. How long would it take $3000 to grow to $ 6000? \n" ); document.write( "
Algebra.Com's Answer #183960 by Edwin McCravy(20060)\"\" \"About 
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A GIC pays 6% per annum. How long would it take $3000 to grow to $ 6000?
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document.write( "\"A+=+P%281%2Br%2Fn%29%5E%28nt%29\"\r\n" );
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document.write( "We solve for t:\r\n" );
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document.write( "\"P%281%2Br%2Fn%29%5E%28nt%29=A\"\r\n" );
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document.write( "Take logs of both sides:\r\n" );
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document.write( "\"log%28%28P%281%2Br%2Fn%29%5E%28nt%29%29%29=log%28%28A%29%29\"\r\n" );
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document.write( "Use a rule of the log of a product:\r\n" );
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document.write( "\"log%28%28P%29%29%2Blog%28%281%2Br%2Fn%29%5E%28nt%29%29=log%28%28A%29%29\"\r\n" );
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document.write( "Subtract \"log%28%28P%29%29\" from both sides:\r\n" );
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document.write( "\"log%28%281%2Br%2Fn%29%5E%28nt%29%29=log%28%28A%29%29-log%28%28P%29%29\"\r\n" );
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document.write( "Use a rule of the log of a power:\r\n" );
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document.write( "\"nt%2Alog%28%281%2Br%2Fn%29%29=log%28%28A%29%29-log%28%28P%29%29\"\r\n" );
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document.write( "Divide both sides by \"n%2Alog%28%281%2Br%2Fn%29%29\"\r\n" );
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document.write( "\"t+=+%28log%28%28A%29%29-log%28%28P%29%29%29%2F%28n%2Alog%28%281%2Br%2Fn%29%29%29\"\r\n" );
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document.write( "In your problem, \r\n" );
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document.write( "P = 3000\r\n" );
document.write( "A = 6000\r\n" );
document.write( "r = 6% = .06\r\n" );
document.write( "n = 1   (per annum)\r\n" );
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document.write( "The 11th year the amount would be \r\n" );
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document.write( "\"A+=+P%281%2Br%2Fn%29%5E%28nt%29\"\r\n" );
document.write( "\"A+=+3000%281%2B.06%2F1%29%5E%281%2A11%29\"\r\n" );
document.write( "\"A+=+%22%245694.89%22\"  [Interest-payers always round down to the lower penny,\r\n" );
document.write( "                       never up to the higher penny]\r\n" );
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document.write( "The 12th year it would be\r\n" );
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document.write( "\"A+=+P%281%2Br%2Fn%29%5E%28nt%29\"\r\n" );
document.write( "\"A+=+3000%281%2B.06%2F1%29%5E%281%2A12%29\"\r\n" );
document.write( "\"A+=+%22%246036.58%22\" \r\n" );
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document.write( "So the 12th year would be the first time it would\r\n" );
document.write( "have been at least $6000.\r\n" );
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document.write( "Edwin
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