document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=72953>Question 72953</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The amount of radioactive tracer remaining after t days is given by A =Ao e-o.o58t, where AO 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( "please show me\r<BR>\n" );
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document.write( "Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #52258 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=52258\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+A%5B0%5D%2Ae%5E%28-0.058%2At%29 ALIGN=MIDDLE  ALT=\"A+=+A%5B0%5D%2Ae%5E%28-0.058%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "A is the amount of material at time t.  <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 the starting amount of material. And<BR>\n" );
document.write( "t is the amount of elapsed time in days. You are asked to find the amount of time t that it<BR>\n" );
document.write( "will take for the amount of material to equal half the amount you started with.<BR>\n" );
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document.write( "Since you started with <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;\" > when it is half gone the A will equal <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%2A+A%5B0%5D ALIGN=MIDDLE  ALT=\"%281%2F2%29%2A+A%5B0%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Substituting this in for A results in the equation becoming:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29%2AA%5B0%5D+=+A%5B0%5D%2Ae%5E%28-0.058%2At%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29%2AA%5B0%5D+=+A%5B0%5D%2Ae%5E%28-0.058%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "If you divide both sides of this equation by <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;\" > the <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;\" > term on both sides<BR>\n" );
document.write( "drops out and you are left with:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29+=+e%5E%28-0.058%2At%29 ALIGN=MIDDLE  ALT=\"%281%2F2%29+=+e%5E%28-0.058%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Take the logarithm of both sides.  Since the equation has an e in it, take the natural logarithm,<BR>\n" );
document.write( "ln:<BR>\n" );
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document.write( "ln(1/2) = ln(e^(-0.058*t))<BR>\n" );
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document.write( "On a calculator you will find that ln(1/2) = ln(0.5) = -0.69314718.  Substitute this into<BR>\n" );
document.write( "the equation for ln(1/2) to get:<BR>\n" );
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document.write( "-0.69314718 = ln(e^(-0.058*t))<BR>\n" );
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document.write( "On the right side, by the rules of logarithms in a logarithm of a term with an exponent<BR>\n" );
document.write( "the exponent can be taken as the multiplier of the logarithm. As a result, the term:<BR>\n" );
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document.write( "ln(e^(-0.058*t))<BR>\n" );
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document.write( "is equal to the term:<BR>\n" );
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document.write( "(-0.058*t)*ln(e)<BR>\n" );
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document.write( "but ln(e) is equal to 1.  So this term further reduces to (-0.058*t).  Substitute this<BR>\n" );
document.write( "into the equation and you now have:<BR>\n" );
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document.write( "-0.69314718 = (-0.058*t)<BR>\n" );
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document.write( "Multiply both sides by -1 to eliminate the negative signs and have:<BR>\n" );
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document.write( "0.69314718 = 0.058*t<BR>\n" );
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document.write( "Finally divide both sides by 0.058 to solve for t:<BR>\n" );
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document.write( "0.69314718/0.058 = t<BR>\n" );
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document.write( "And after the division is performed the equation becomes:<BR>\n" );
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document.write( "11.9508 = t<BR>\n" );
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document.write( "So after about 12 days one-half of the original material has decayed.<BR>\n" );
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document.write( "Hope this helps you to decipher the problem. </SPAN>\n" );
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