document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1062936>Question 1062936</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A scientist has 200 grams of a substance where half of it decays every hour. Write an equation to determine how long it takes until 3.125 grams are remaining </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #677943 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=677943\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f = p * e^(rt)\r<BR>\n" );
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document.write( "this is the formula for continuous growth or decay.\r<BR>\n" );
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document.write( "if r is negative, then decay.<BR>\n" );
document.write( "if r is positive, then growth.\r<BR>\n" );
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document.write( "f = future value<BR>\n" );
document.write( "p = present value<BR>\n" );
document.write( "e = scientific constant of 2.718281828.....<BR>\n" );
document.write( "r = rate per time period<BR>\n" );
document.write( "t = number of time periods\r<BR>\n" );
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document.write( "you are given that p = 200<BR>\n" );
document.write( "you are given that f = 100<BR>\n" );
document.write( "you are given that t = 1 hour<BR>\n" );
document.write( "you need to solve for r.\r<BR>\n" );
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document.write( "formula of f = p * e^(rt) becomes 100 = 200 * e^(r*1) which becomes 100 = 200 * e^(r)\r<BR>\n" );
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document.write( "divide both sides of this equation by 200 to get .5 = e^r\r<BR>\n" );
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document.write( "take the natural log of both sides of this equation to get ln(.5) = ln(e^r).\r<BR>\n" );
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document.write( "this is equivalent to ln(.5) = r * ln(e).\r<BR>\n" );
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document.write( "this is equivalent to ln(.5) = r because ln(e) is equal to 1.\r<BR>\n" );
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document.write( "solve for r to get r = ln(.5) = -.6931471806 per hour.\r<BR>\n" );
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document.write( "you started with 100 = 200 * e^r.\r<BR>\n" );
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document.write( "replace r with -.6931471806 to get 100 = 200 * e^-.6931471806.\r<BR>\n" );
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document.write( "simplify to get 100 = 100.\r<BR>\n" );
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document.write( "this confirms the solution for r is correct.\r<BR>\n" );
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document.write( "now that you know r, you want to find the number of hours for the substance to become equal to 3.125 grams.\r<BR>\n" );
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document.write( "same formula of f = p * e^(rt) is used, only this time we are looking for t.\r<BR>\n" );
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document.write( "formula becomes 3.125 = 200 * e^(-.6931471806 * t)\r<BR>\n" );
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document.write( "divide both sides of the equation by 200 to get 3.125/200 = e^(-.6931471806 * t)\r<BR>\n" );
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document.write( "take the natural log of both sides of the equation to get ln(3.125/200) = ln(e^(-.6931471806 * t))\r<BR>\n" );
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document.write( "this is equivalent to ln(3.125/200) = -.6931471806 * t * ln(e)\r<BR>\n" );
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document.write( "since ln(e) = 1, this becomes ln(3.125/200) = -.6931471806 * t\r<BR>\n" );
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document.write( "divide both sides of the equation by -.6931471806 and solve for t to get t = ln(3.125/200) / -.6931471806.\r<BR>\n" );
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document.write( "this makes t equal to 6 hours.\r<BR>\n" );
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document.write( "the mass of 200 grams will reduce to 3.125 grams in 6 hours.\r<BR>\n" );
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document.write( "formula becomes f = 200 * e^(-.6931471806 * 6) which results in f = 3.125 grams.\r<BR>\n" );
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document.write( "this was solved based on continuous compounding.\r<BR>\n" );
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document.write( "you could also have solved using discrete compounding.\r<BR>\n" );
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document.write( "the formula, in that case, would be f = p * (1+r)^n\r<BR>\n" );
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document.write( "f = 100<BR>\n" );
document.write( "p = 200<BR>\n" );
document.write( "n = 1\r<BR>\n" );
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document.write( "the time, in this case, is represented by n.<BR>\n" );
document.write( "it doesn't really matter what variable name you use, as long as you are consistent within the confines of the formula you are using.\r<BR>\n" );
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document.write( "formula becomes 100 = 200 * (1+r)^1 which becomes 100 = 200 * (1+r).\r<BR>\n" );
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document.write( "divide both sides of the equation by 200 and simplify to get .5 = 1 + r\r<BR>\n" );
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document.write( "subtract 1 from both sides of the equation and solve for r to get r = -.5 per hour.\r<BR>\n" );
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document.write( "this says that the substance becomes half of what it was every hour.\r<BR>\n" );
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document.write( "now you need to use the same formula to solve for n.\r<BR>\n" );
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document.write( "formula of f = p * (1 + r)^n becomes 3.125 = 200 * (1 - .5)^n.\r<BR>\n" );
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document.write( "simplify to get 3.125 = 200 * .5^n\r<BR>\n" );
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document.write( "divide both sides of the equation by 200 to get 3.125/200 = .5^n.\r<BR>\n" );
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document.write( "take the log of both sides to get log(3.125/200) = log(.5^n).\r<BR>\n" );
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document.write( "this is equivalent to log(3.125/200) = n * log(.5)\r<BR>\n" );
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document.write( "divide both sides by log(.5) and solve for n to get n = log(3.125/200)/log(.5) = 6.\r<BR>\n" );
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document.write( "the substance will reduce from 200 grams to 3.125 grams in 6 hours.\r<BR>\n" );
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document.write( "you get the same answer whether you used continuous compounding formula or discrete compounding formula.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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