document.write( "Question 4186: Okay. I know I asked this already, and I thank you longjohnsilver for your post; but quite frankly, that makes no sense. I need some kind of basic formula for annual interest. I remember learning it, but can't recall it all.\r
\n" ); document.write( "\n" ); document.write( "A BOND PAYS 12% ANNUAL INTEREST COMPOUNDED CONTINUOUSLY. IF $10,000 IS INVESTED INITIALLY, WHEN WILL THE BOND BE WORTH $30,000? ROUND TO NEAREST YEAR.
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Algebra.Com's Answer #1866 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
OK, try this:\r
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\n" ); document.write( "\n" ); document.write( "start with 10000.
\n" ); document.write( "After 1 year you have 10000*1.12 = whatever.
\n" ); document.write( "After another year, you have whatever*1.12 = something.
\n" ); document.write( "After a third year, you have something*1.12 = value
\n" ); document.write( "After a fourth year, you have a value*1.12 = figure
\n" ); document.write( "etc etc.\r
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\n" ); document.write( "\n" ); document.write( "Work that out until you reach 30,000. It will be 10 years.\r
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\n" ); document.write( "\n" ); document.write( "Mathematically, what we have is:\r
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\n" ); document.write( "\n" ); document.write( "after year 1: 10000*1.12
\n" ); document.write( "after year 2: 10000*1.12*1.12
\n" ); document.write( "after year 3: 10000*1.12*1.12*1.12
\n" ); document.write( "after year 4: 10000*1.12*1.12*1.12*1.12
\n" ); document.write( "after year 5: 10000*1.12*1.12*1.12*1.12*1.12
\n" ); document.write( "etc\r
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\n" ); document.write( "\n" ); document.write( "which can be written as:\r
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\n" ); document.write( "\n" ); document.write( "after 1 year : 10000*1.12^1
\n" ); document.write( "after 2 years: 10000*1.12^2
\n" ); document.write( "after 3 years: 10000*1.12^3
\n" ); document.write( "after 4 years: 10000*1.12^4
\n" ); document.write( "after 5 years: 10000*1.12^5\r
\n" ); document.write( "\n" ); document.write( "... after n years: 10000*1.12^n and this has to be 30000. So:\r
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\n" ); document.write( "\n" ); document.write( "\"10000%2A1.12%5En+=+30000\"\r
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\n" ); document.write( "\n" ); document.write( "\"1.12%5En+=+3\". Taking logs:\r
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\n" ); document.write( "\n" ); document.write( "nlog1.12 = log3
\n" ); document.write( "n=log3/log1.12
\n" ); document.write( "n=9.694\r
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\n" ); document.write( "\n" ); document.write( "so n=10 years\r
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\n" ); document.write( "\n" ); document.write( "jon
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