document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=759722>Question 759722</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The normal healing of wounds can be modeled by an exponential function. If A0 represents the original area of the wound and if A equals the area of the would after n days, then the formula below describes the area of a wound on the nth day following an injury when no infection is present to retard the healing.\r<BR>\n" );
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document.write( "    A = A_0e^-0.36n \r<BR>\n" );
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document.write( "Suppose that a wound initially had an area of 100 square millimeters.\r<BR>\n" );
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document.write( "1.If healing is taking place, how long will it be before the wound is exactly one-half its original size?<BR>\n" );
document.write( "2.How long will it be before the wound is exactly 10% of its original size?\r<BR>\n" );
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document.write( "I cant figure out how to calculate this i dont understand A_0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #462207 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><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=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=462207\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> hi there,<BR>\n" );
document.write( "Normally the equation would be laid out like this:<BR>\n" );
document.write( "A(t) = A(o)e^-0.36t<BR>\n" );
document.write( "A(t) = time after the event has taken place.<BR>\n" );
document.write( "A(o) = time at the start of the event.\r<BR>\n" );
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document.write( "1) How long before wound is half its size<BR>\n" );
document.write( "  A(t) = A(o)e^-0.36t<BR>\n" );
document.write( "  50 = 100 e^-0.36t<BR>\n" );
document.write( "  50/100 = e^-0.36t<BR>\n" );
document.write( "    0.5 = e^-0.36t<BR>\n" );
document.write( "Take loge of both sides.<BR>\n" );
document.write( "-0.36t comes in front<BR>\n" );
document.write( "  loge 0.5 = -0.36t loge e  (loge e = 1)<BR>\n" );
document.write( "  loge 0.5 = -0.36t<BR>\n" );
document.write( "loge 0.5/-0.36 = t<BR>\n" );
document.write( "         t = 1.9 or 2 days.<BR>\n" );
document.write( "2) 10% of its original size<BR>\n" );
document.write( "10% of 100 = 10<BR>\n" );
document.write( "A(t) = A(o)e^-0.36t<BR>\n" );
document.write( "10 = 100e^-0.36t<BR>\n" );
document.write( "10/100 = e^-0.36t<BR>\n" );
document.write( "  0.1 = e^-0.36t<BR>\n" );
document.write( "Take loge of both sides<BR>\n" );
document.write( "-0.36t comes in front<BR>\n" );
document.write( "loge 0.1 = -0.36t loge e<BR>\n" );
document.write( "  loge 0.1/-0.36 = t<BR>\n" );
document.write( "            t = 6.39days.<BR>\n" );
document.write( "Hope this helps.<BR>\n" );
document.write( ":-) </SPAN>\n" );
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