document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=898557>Question 898557</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Help with this log problem. I have the answer but am confused on how to get to it\r<BR>\n" );
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document.write( "Prehistoric cave paintings were discovered in a cave in France. The paint contained 39% of the original carbon-14. Use the exponential decay model for carbon-14, A = Ao e ^(-0.001211)(t) to estimate the age of the paintings.\r<BR>\n" );
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document.write( "The paitings are approximately 7.782 years old.\r<BR>\n" );
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document.write( "Can someone please show me how to arrive at that answer?\r<BR>\n" );
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document.write( "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 #544895 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39620)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=544895\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your decay equation is A=Ao*e^(-0.001211t) in pure text and using your same variables.  This will appear more conventionally as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=Ao%2Ae%5E%28-0.001211t%29 ALIGN=MIDDLE  ALT=\"A=Ao%2Ae%5E%28-0.001211t%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "You are given a relationship between A and Ao, and are asked for t.  Solve first symbolically for t, which is a more comfortable process.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28A%29=ln%28Ao%29%2Bln%28e%5E%28-0.001211t%29%29 ALIGN=MIDDLE  ALT=\"ln%28A%29=ln%28Ao%29%2Bln%28e%5E%28-0.001211t%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28A%29=ln%28Ao%29%2B%28-0.001211t%29%2A1 ALIGN=MIDDLE  ALT=\"ln%28A%29=ln%28Ao%29%2B%28-0.001211t%29%2A1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28A%29-ln%28Ao%29=-0.001211t ALIGN=MIDDLE  ALT=\"ln%28A%29-ln%28Ao%29=-0.001211t\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28Ao%29-ln%28A%29=0.001211t ALIGN=MIDDLE  ALT=\"ln%28Ao%29-ln%28A%29=0.001211t\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%28t=%281%2F0.001211%29ln%28Ao%2FA%29%29 ALIGN=MIDDLE  ALT=\"highlight_green%28t=%281%2F0.001211%29ln%28Ao%2FA%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "You were given A is a percentage of Ao, so take Ao as 100 and A as 39.<BR>\n" );
document.write( "Now just substitute into the formula for t:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=t=%281%2F0.001211%29ln%28100%2F39%29 ALIGN=MIDDLE  ALT=\"t=%281%2F0.001211%29ln%28100%2F39%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "...with a calculator for help....<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28t=778%29 ALIGN=MIDDLE  ALT=\"highlight%28t=778%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > years.\r<BR>\n" );
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document.write( "This showed at four significant figures to be 777.5 years, but how accurately you can trust the decay equation can be debated.  You most likely made a decimal mistake in your steps. </SPAN>\n" );
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