document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=73053>Question 73053</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have only 3 questions like this\r<BR>\n" );
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document.write( "The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?\r<BR>\n" );
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document.write( "47.9\r<BR>\n" );
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document.write( "48.8\r<BR>\n" );
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document.write( "61.6\r<BR>\n" );
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document.write( "68.6\r<BR>\n" );
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document.write( "please teach me through this. Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #52317 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=52317\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The amount of a radioactive tracer remaining after t days is given by A = AO e-0.18t, where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?<BR>\n" );
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document.write( "Radio active decay formula should be given as:<BR>\n" );
document.write( "A = Ao(e^-.18t)<BR>\n" );
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document.write( "A = 40; t = 3; Find Ao<BR>\n" );
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document.write( "40 = Ao(e^-.18*3)<BR>\n" );
document.write( "40 = Ao(e^-.54<BR>\n" );
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document.write( "Using a good calculator; enter e^(-.54), should equal .5827482524<BR>\n" );
document.write( "So you have:<BR>\n" );
document.write( ".5827482524Ao = 40<BR>\n" );
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document.write( "Ao = 40/.5827482524<BR>\n" );
document.write( "Ao = 68.64 grams to be acquired<BR>\n" );
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document.write( "If they want you to use natural logs, you add logs when you multiply:<BR>\n" );
document.write( "ln(40) = ln(Ao) + ln(e^-.54)<BR>\n" );
document.write( "ln(40) = ln(Ao) + (-.54)(ln(e), log equiv of exponents<BR>\n" );
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document.write( "Find the ln of 40. Remember the ln of e is 1, so we have:<BR>\n" );
document.write( "3.688879454 = ln(Ao) - .54<BR>\n" );
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document.write( "3.688879454 +.54 = ln(Ao)<BR>\n" );
document.write( "4.228879454 = ln(Ao)<BR>\n" );
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document.write( "Find the e^x; enter e^4.228879454:<BR>\n" );
document.write( "Ao = 68.64, the same answer\r<BR>\n" );
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document.write( " </SPAN>\n" );
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