document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1201494>Question 1201494</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I know that we are not supposed to put 2 questions in 1 however, they both go together so I felt as though it was best to put them both on one submission. \r<BR>\n" );
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document.write( "The radioactive isotope of potassium-42, which is vital in the diagnosis of brain tumors, has a half life of 12.36 hours.\r<BR>\n" );
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document.write( "Determine the exponential decay model that represents the mass of potassium-42.If 500-mg of potassium-42 was taken, how much milligrams of this isotope will remain after 48 hours?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #835884 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=835884\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "x = number of hours that elapse<BR>\n" );
document.write( "y = amount, in mg, of potassium-42 remaining<BR>\n" );
document.write( "H = half-life of potassium-42 = 12.36 hours\r<BR>\n" );
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document.write( "Half-life template equation<BR>\n" );
document.write( "y = a*(0.5)^(x/H)\r<BR>\n" );
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document.write( "We could use 1/2 in place of 0.5, but I find the 0.5 is easier to work with here.\r<BR>\n" );
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document.write( "Let's walk through a few scenarios for x.<ul><li>If one half-life elapses, then x = H. The exponent x/H becomes H/H = 1, meaning we have one copy of 0.5 multiplied to the starting amount 'a'. Half of the substance remains.</li><li>If x = 2H, two half-lives elapse, and x/H becomes 2H/H = 2. We'll have 2 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^2 = 0.25 = 25% of the substance remains. </li><li>If x = 3H, three half-lives elapse, and x/H becomes 3H/H = 3. We'll have 3 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^3 = 0.125 = 12.5% of the substance remains. </li></ul>And so on.\r<BR>\n" );
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document.write( "y = a*(0.5)^(x/H)<BR>\n" );
document.write( "y = 500*(0.5)^(x/12.36)<BR>\n" );
document.write( "This is one way to represent the exponential model, aka equation, aka function.\r<BR>\n" );
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document.write( "We have exponential decay because the base b = 0.5 fits the interval 0 < b < 1.\r<BR>\n" );
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document.write( "There are other formats to express this equation into.<BR>\n" );
document.write( "Here are the steps to arrive at a second format.<BR>\n" );
document.write( "y = 500*(0.5)^(x/12.36)<BR>\n" );
document.write( "y = 500 * [ (0.5)^(1/12.36) ]^x<BR>\n" );
document.write( "y = 500 * ( 0.94546361966231 )^x<BR>\n" );
document.write( "The decimal value is approximate.<BR>\n" );
document.write( "Round this value however your teacher instructs.\r<BR>\n" );
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document.write( "Here's yet another pathway to determine an equation.<BR>\n" );
document.write( "Compare y = 500 * ( 0.94546361966231 )^x with y = 500*e^(kx) aka y = 500*(e^k)^x<BR>\n" );
document.write( "We see that<BR>\n" );
document.write( "e^k = 0.94546361966231<BR>\n" );
document.write( "k = Ln(0.94546361966231)<BR>\n" );
document.write( "k = -0.05607986897736<BR>\n" );
document.write( "Therefore,<BR>\n" );
document.write( "y = 500*e^(kx)<BR>\n" );
document.write( "y = 500*e^(-0.05607986897736x)\r<BR>\n" );
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document.write( "So we have this candidate list of choices for the equation<ul><li>y = 500*(0.5)^(x/12.36)</li><li>y = 500 * ( 0.94546361966231 )^x</li><li>y = 500*e^(-0.05607986897736x)</li></ul>There are probably other formats to consider.\r<BR>\n" );
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document.write( "Personally I prefer the first format because I think it's easiest to work with. <BR>\n" );
document.write( "The half-life H = 12.36 is readily visible without having to do any calculations.\r<BR>\n" );
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document.write( "After we've determined the equation, we plug in x = 48 to figure out how much potassium-42 is left after 48 hours.<BR>\n" );
document.write( "y = 500*(0.5)^(x/12.36)<BR>\n" );
document.write( "y = 500*(0.5)^(48/12.36)<BR>\n" );
document.write( "y = 33.8782897037247<BR>\n" );
document.write( "y = 33.88<BR>\n" );
document.write( "or you could say<BR>\n" );
document.write( "y = 500 * ( 0.94546361966231 )^x<BR>\n" );
document.write( "y = 500 * ( 0.94546361966231 )^48<BR>\n" );
document.write( "y = 33.8782897037107<BR>\n" );
document.write( "y = 33.88<BR>\n" );
document.write( "or<BR>\n" );
document.write( "y = 500*e^(-0.05607986897736x)<BR>\n" );
document.write( "y = 500*e^(-0.05607986897736*48)<BR>\n" );
document.write( "y = 33.8782897036631<BR>\n" );
document.write( "y = 33.88<BR>\n" );
document.write( "About 33.88 mg of potassium-42 remains after 48 hours.\r<BR>\n" );
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document.write( "An informal way to think about it:<BR>\n" );
document.write( "1 half-life = 12.36 hrs<BR>\n" );
document.write( "1 hr = 1/(12.36) half-lives <BR>\n" );
document.write( "48 hr = 48/(12.36) half-lives <BR>\n" );
document.write( "48 hr = 3.8835 half-lives approximately<BR>\n" );
document.write( "The answer will involve a duration between 3 and 4 half-lives.\r<BR>\n" );
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document.write( "We start with 500 mg of potassium-42.<ul><li>After 12.36 hours (1 half-life), half of it goes away. We have 0.5*500 = 250 mg left.</li><li>After 24.72 hours (2 half-lives = 2*12.36 = 24.72 hrs), there will be 0.5*250 = 125 mg left.</li><li>After 37.08 hours (3 half-lives = 3*12.36 = 37.08 hrs), there will be 0.5*125 = 62.5 mg left.</li><li>After 49.44 hours (4 half-lives = 4*12.36 = 49.44 hrs), there will be 0.5*62.5 = 31.25 mg left.</li></ul>Therefore, the answer must be between 62.5 mg and 31.25 mg (corresponding to 3 and 4 half-lives).<BR>\n" );
document.write( "This method won't nail down the exact value, but it does give an informal way to figure out where the answer should be located. <BR>\n" );
document.write( "It can serve as a way to check the answer.\r<BR>\n" );
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document.write( "<font color=red>Summary:</font>\r<BR>\n" );
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document.write( "One equation is <font color=red>y = 500*(0.5)^(x/12.36)</font><BR>\n" );
document.write( "Other equations are possible. See above.\r<BR>\n" );
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document.write( "x = number of hours that elapse<BR>\n" );
document.write( "y = amount, in mg, of potassium-42 remaining\r<BR>\n" );
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document.write( "<font color=red>Approximately 33.88 mg</font> of potassium-42 remains after 48 hours. <BR>\n" );
document.write( "Round this value however your teacher instructs.<BR>\n" );
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