document.write( "Question 244452: If an investment of $8000 increases by 15 percent at the end of each year, what is the fewest number of years until it doubles in value?
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Algebra.Com's Answer #178787 by Edwin McCravy(20056)\"\" \"About 
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If an investment of $8000 increases by 15 percent at the end of each year, what is the fewest number of years until it doubles in value?
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document.write( "If it increases by 15% it becomes 115% of what it was before. To take 115% of \r\n" );
document.write( "something is to multiply it by 1.15.  So we want to know how many times\r\n" );
document.write( "we must multiply $8000 by 1.15 until it just becomes $16000, or first exceeds\r\n" );
document.write( "it.  Let that number be n.  Then to get the approximate nu,ber of years\r\n" );
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document.write( "\"8000%2A1.15%5En+=+16000\"\r\n" );
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document.write( "Take the log of both sides:\r\n" );
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document.write( "\"log%288000%2A1.15%5En%29+=+log%2816000%29\"\r\n" );
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document.write( "\"log%288000%29%2Blog%281.15%5En%29+=+log%2816000%29\"\r\n" );
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document.write( "\"log%281.15%5En%29+=+log%2816000%29-log%288000%29\"\r\n" );
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document.write( "\"n%2Alog%281.15%29+=+log%2816000%29-log%288000%29\"\r\n" );
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document.write( "Divide both sides by \"log%281.15%29\"\r\n" );
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document.write( "\"n+=+%28log%2816000%29-log%288000%29%29%2Flog%281.15%29\"\r\n" );
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document.write( "\"n+=+4.959484455\"\r\n" );
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document.write( "or in 4 years it would be under $16000 and in 5 years\r\n" );
document.write( "it would be over $16000.\r\n" );
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document.write( "Edwin
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