document.write( "Question 1130906: Ten thousand dollars is deposited in an account that pays 6% annual interest. How long does it take to double the initial deposit if the interest is compounded annually? monthly? daily? \n" ); document.write( "
Algebra.Com's Answer #747531 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The doubling time is independent of the initial amount.

\n" ); document.write( "At 6% annual interest for t years, the growth factor over the initial amount is

\n" ); document.write( "\"%281%2B%28.06%2Fn%29%29%5E%28n%2At%29\" where n is the number of compounding periods per year.

\n" ); document.write( "To find the doubling time, we want the growth factor to be equal to 2. So

\n" ); document.write( "(a) for annual compounding, the number of periods per year is 1:

\n" ); document.write( "\"%281%2B%28.06%2F1%29%29%5E%28%281%2At%29%29+=+2\"

\n" ); document.write( "(b) for monthly compounding, the number of periods per year is 12:

\n" ); document.write( "\"%281%2B%28.06%2F12%29%29%5E%28%2812%2At%29%29+=+2\"

\n" ); document.write( "(c) for daily compounding, it is customary to use 360 as the number of periods per year:

\n" ); document.write( "\"%281%2B%28.06%2F360%29%29%5E%28%28360%2At%29%29+=+2\"

\n" ); document.write( "The unknown (number of years, t) is an exponent in each of these equations; to solve each equation you need to use logarithms, or a graphing calculator. Here is a solution for monthly compounding using logarithms.

\n" ); document.write( "\"%281%2B%28.06%2F12%29%29%5E%28%2812%2At%29%29+=+2\"

\n" ); document.write( "\"%281.005%29%5E%28%2812%2At%29%29+=+2\"
\n" ); document.write( "\"12t%2Alog%281.005%29+=+log%282%29\"
\n" ); document.write( "\"t+=+log%282%29%2F%2812%2Alog%281.005%29%29\" = 11.58 years, to 2 decimal places

\n" ); document.write( "You can work the cases for annual and daily compounding. The daily compounding should take less time than the monthly to double, and the annual compounding should take longer then the monthly to double.
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