document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=48240>Question 48240</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0534373526/algebrahomeworkh>College Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I worked this problem out but would like to check with you if I'm vaguely correct: <BR>\n" );
document.write( "The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample.<BR>\n" );
document.write( "a. Find a function that models the mass remaining after t years\r<BR>\n" );
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document.write( "Using the model for radioactive decay mt = m0e^-rt:<BR>\n" );
document.write( "m0 = 22mg, r=(ln2/1600) = -0.00043\r<BR>\n" );
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document.write( "m(t) = 22e^-0.00043t\r<BR>\n" );
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document.write( "b. How much of the sample will remain after 4000 years?\r<BR>\n" );
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document.write( "t = 4000<BR>\n" );
document.write( "m(4000) = 22e^(-0.00043)(4000) = 3.94 mg\r<BR>\n" );
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document.write( "c. After how long will only 18 mg of the sample remain?\r<BR>\n" );
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document.write( "m(t) = 18\r<BR>\n" );
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document.write( "22e^-0.00043t = 18<BR>\n" );
document.write( "e^-0.00043t = 9/11<BR>\n" );
document.write( "Ln e^-0.00043t = ln 9/11<BR>\n" );
document.write( "0.00043t = ln 9/11<BR>\n" );
document.write( "t = -(ln(9/11)/0.0043) = 46.7 years\r<BR>\n" );
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document.write( "Thanks a million(factored many times) for your help! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #31911 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=31911\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample.<BR>\n" );
document.write( "a. Find a function that models the mass remaining after t years <BR>\n" );
document.write( "Using the model for radioactive decay mt = m0e^-rt:<BR>\n" );
document.write( "m0 = 22mg, r=(ln2/1600) = -0.00043<BR>\n" );
document.write( "-----------------------------------<BR>\n" );
document.write( "A rate cannot be negative.<BR>\n" );
document.write( "You probably mean r=0.00043 <BR>\n" );
document.write( "----------------------------<BR>\n" );
document.write( "m(t) = 22e^-0.00043t <BR>\n" );
document.write( "b. How much of the sample will remain after 4000 years? <BR>\n" );
document.write( "t = 4000<BR>\n" );
document.write( "m(4000) = 22e^(-0.00043)(4000) = 3.94 mg <BR>\n" );
document.write( "c. After how long will only 18 mg of the sample remain? <BR>\n" );
document.write( "m(t) = 18 <BR>\n" );
document.write( "22e^-0.00043t = 18<BR>\n" );
document.write( "e^-0.00043t = 9/11<BR>\n" );
document.write( "Ln e^-0.00043t = ln 9/11<BR>\n" );
document.write( "0.00043t = ln 9/11<BR>\n" );
document.write( "-----------------------<BR>\n" );
document.write( "I think you need the following:<BR>\n" );
document.write( "ln e^-0.00043t = ln 9/11<BR>\n" );
document.write( "-0.00043t = ln(9/11)<BR>\n" );
document.write( "t=[ln(9/11)]/(-0.00043)<BR>\n" );
document.write( "t=466.676 years<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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