document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=80072>Question 80072</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0534494498/algebrahomeworkh>Algebra and Trigonometry with Analytic Geometry</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> PROBLEM:\r<BR>\n" );
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document.write( "The amount of a radioactive tracer remaining after 't' days is given by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=A0e%5E%28-0.058t%29 ALIGN=MIDDLE  ALT=\"A=A0e%5E%28-0.058t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, where A0 is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay?\r<BR>\n" );
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document.write( "OPTIONS:<BR>\n" );
document.write( "a. 10 days<BR>\n" );
document.write( "b. 11 days<BR>\n" );
document.write( "c. 12 days<BR>\n" );
document.write( "d. 13 days\r<BR>\n" );
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document.write( "MY WORK SO FAR:\r<BR>\n" );
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document.write( "FORMULA: \"Law of growth or decay\" let 'A0' be the value of a quantity 'A' at time 't'=0 (that is, 'A0' is the initial amount of 'A'.) If 'A' changes instantaneously at a rate proportional to itscurrent value, then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=A%28t%29=A0e%5Ert ALIGN=MIDDLE  ALT=\"A=A%28t%29=A0e%5Ert\"   DISABLED_style= \"max-width:90%; height:auto;\" > where r>0 is the rate of growth (or r<0 is the rate of decay) of A.\r<BR>\n" );
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document.write( "therefore if we use the formula then we have \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=A0%282.71828%29%5E%28-0.058t%29 ALIGN=MIDDLE  ALT=\"A=A0%282.71828%29%5E%28-0.058t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > \r<BR>\n" );
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document.write( "QUESTIONS:\r<BR>\n" );
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document.write( "1. What is the first step into solving this problem? <BR>\n" );
document.write( "2. What are the ways to gather the information needed to continue solving this problem? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #57427 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=57427\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=A%5B0%5De%5E%28-0.058t%29 ALIGN=MIDDLE  ALT=\"A=A%5B0%5De%5E%28-0.058t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Let's just assume that the original amount <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%5B0%5D ALIGN=MIDDLE  ALT=\"A%5B0%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > is 1.  Then to decay by half the amount<BR>\n" );
document.write( "the resulting value of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A ALIGN=MIDDLE  ALT=\"A\"   DISABLED_style= \"max-width:90%; height:auto;\" > would be <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Substitute these two values and you<BR>\n" );
document.write( "get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2=1%2Ae%5E%28-0.058t%29 ALIGN=MIDDLE  ALT=\"1%2F2=1%2Ae%5E%28-0.058t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "and after the multiplication by 1, the right side becomes just:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2=e%5E%28-0.058t%29 ALIGN=MIDDLE  ALT=\"1%2F2=e%5E%28-0.058t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Because an exponent contains the variable we are to solve for, that's a clue that we can use<BR>\n" );
document.write( "logarithms and the rules of logarithms to solve.  And because e is also involved, let's<BR>\n" );
document.write( "take the natural log of both sides to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%281%2F2%29+=+ln%28e%5E%28-0.058t%29%29 ALIGN=MIDDLE  ALT=\"ln%281%2F2%29+=+ln%28e%5E%28-0.058t%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "A rule of logarithms that can be applied is that if the quantity that the logarithm<BR>\n" );
document.write( "acts on has an exponent, you can make that exponent a multiplier of the logarithm.<BR>\n" );
document.write( "Applying this rule to our problem results in:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%281%2F2%29+=+-0.058t%2Aln%28e%29 ALIGN=MIDDLE  ALT=\"ln%281%2F2%29+=+-0.058t%2Aln%28e%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "But the natural logarithm of e is equal to 1.  So substitute 1 for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28e%29 ALIGN=MIDDLE  ALT=\"ln%28e%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%281%2F2%29+=+-0.058t+%2A+1 ALIGN=MIDDLE  ALT=\"ln%281%2F2%29+=+-0.058t+%2A+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Simplify the right side by doing the multiplication and get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%281%2F2%29+=+-0.058t ALIGN=MIDDLE  ALT=\"ln%281%2F2%29+=+-0.058t\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now you can use a calculator to find the natural logarithm of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F2 ALIGN=MIDDLE  ALT=\"1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > or its equivalent 0.5.<BR>\n" );
document.write( "This logarithm is -0.69314718 and when this is substituted for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%281%2F2%29 ALIGN=MIDDLE  ALT=\"ln%281%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > the equation<BR>\n" );
document.write( "becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+-0.69314718+=+-0.058t ALIGN=MIDDLE  ALT=\"+-0.69314718+=+-0.058t\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Finally solve for t by dividing both sides by -0.058 to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=t+=+%28-0.69314718%29%2F%28-0.058%29+=+11.95081346 ALIGN=MIDDLE  ALT=\"t+=+%28-0.69314718%29%2F%28-0.058%29+=+11.95081346\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Therefore, the closest answer in your list of possible answers is 12 days.  Hope this helps<BR>\n" );
document.write( "you to see a way to solve problems such as these. </SPAN>\n" );
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