document.write( "Question 481976: Determine how much time is required to triple your money if interest is 5.75% and is compounded daily?
\n" ); document.write( "I have to use the compound interest formula: A=P(1+(r/n))^nt
\n" ); document.write( "Thank you, in advance! \n" ); document.write( "
Algebra.Com's Answer #330034 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
A = 3
\n" ); document.write( "P = 1
\n" ); document.write( "r = .0575
\n" ); document.write( "n = 365
\n" ); document.write( "t is what you're solving for.
\n" ); document.write( "your equation becomes:
\n" ); document.write( "3 = 1 * (1 + (1+(.0575/365))^(365*t)
\n" ); document.write( "this equation becomes:
\n" ); document.write( "3 = (1+(.0575/365))^(365*t)
\n" ); document.write( "take the log of both sides of this equation to get:
\n" ); document.write( "log(3) = log(1+(.0575/365))^(365*t)
\n" ); document.write( "since log(a^b) = b*log(a), your equation becomes:
\n" ); document.write( "log(3) = 365*t*log(1+(.0575/365))
\n" ); document.write( "divide both sides of this equation by log(1+(.0575/365)) to get:
\n" ); document.write( "log(3)/log(1+(.0575/365)) = 365*t
\n" ); document.write( "divide both sides of this equation by 365 to get:
\n" ); document.write( "log(3)/(log(1+(.0575/365))*365) = t
\n" ); document.write( "solve for t using your calculator LOG function to get:
\n" ); document.write( "19.1078055813139
\n" ); document.write( "that's a little over 19 years.
\n" ); document.write( "plug that value into your original equation and solve.
\n" ); document.write( "original equation is:
\n" ); document.write( "3 = 1 * (1 + (.0575/365)^(365*t) which becomes:
\n" ); document.write( "3 = 1 * (1 + (.0575/365)^(365*19.1078055813139) which becomes:
\n" ); document.write( "3 = 3.00000000000001
\n" ); document.write( "plug that into your calculator to get:
\n" ); document.write( "3 = 3.00000000000001 which is close enough to say that it's right on.
\n" ); document.write( "the only reason it's not is because the calculator i am using goes very far out so a very tiny rounding error can actually be shown. most calculators wouldn't even show that.
\n" ); document.write( "note:
\n" ); document.write( "there are not 365 days in every year.
\n" ); document.write( "there are 366 days every fourth year.
\n" ); document.write( "you can fine tune your answer by taking 3 years at 365 days each and adding a fourth year at 366 days and then dividing the sum by 4 to get an average of 365.25 days per year.
\n" ); document.write( "your answer might be slightly different but not significantly so.
\n" ); document.write( "the actual answer would be:
\n" ); document.write( "log(3)/(log(1+(.0575/365.25))*365.25) = t
\n" ); document.write( "t = 19.1078045512825 instead of t = 19.10780455
\n" ); document.write( "that's a factor of 19.1078045512825/19.10780455 = 1.00000000006712 over the answer using 365 days a year.
\n" ); document.write( "bottom line is they're so close that it doesn't really matter, for practical purposes, whether you are using 365 days a year or 365.25 days a year.
\n" ); document.write( "either way, your answer is 19.1078 years if you round to the nearest 10,000th of a year.\r
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