document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=117972>Question 117972</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This word problem is very confusing, help me find out the answer:<BR>\n" );
document.write( "A certain radioactive isotope has a half-life of approximately 1900 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 80% of that amount. <BR>\n" );
document.write( "A. 612 years<BR>\n" );
document.write( "B. 380 years<BR>\n" );
document.write( "C. 532 years<BR>\n" );
document.write( "D. 4412 years\r<BR>\n" );
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document.write( "I have tried: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1900=e%5E%28.8x%29 ALIGN=MIDDLE  ALT=\"1900=e%5E%28.8x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Nothing seems to work. Also I know I have to round and use a T.I. and I did, but my answer doesn't come close to the answer choices.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #86008 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=86008\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A certain radioactive isotope has a half-life of approximately 1900 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 80% of that amount.<BR>\n" );
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document.write( "The half-life equation that I am familiar with:<BR>\n" );
document.write( "A = Ao(2^(-t/h))<BR>\n" );
document.write( "where:<BR>\n" );
document.write( "A = resulting amt<BR>\n" );
document.write( "Ao = initial amt<BR>\n" );
document.write( "t = time<BR>\n" );
document.write( "h = half life of the substance<BR>\n" );
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document.write( "Let our Ao = 1 and A = .8<BR>\n" );
document.write( "h = given as 1900 yrs<BR>\n" );
document.write( "find t<BR>\n" );
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document.write( "1(2^(-t/1900)) = .8<BR>\n" );
document.write( "Find the nat log of both sides<BR>\n" );
document.write( "ln(2^-t/1900)) = ln(.8)<BR>\n" );
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document.write( "Log equivalent of exponents<BR>\n" );
document.write( "(-t/1900)*ln(2) = ln(.8)<BR>\n" );
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document.write( "(-t/1900)*.693 = -.223<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F1900 ALIGN=MIDDLE  ALT=\"-t%2F1900\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-.223%2F.693 ALIGN=MIDDLE  ALT=\"-.223%2F.693\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F1900 ALIGN=MIDDLE  ALT=\"-t%2F1900\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -.322<BR>\n" );
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document.write( "Multiply both sides by -1900<BR>\n" );
document.write( "t = -1900 * -.322<BR>\n" );
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document.write( "t = 611.8 years<BR>\n" );
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document.write( "You can check this on your calc: enter: 2^(-611.8/1900) = .79996 = .8 or 80%<BR>\n" );
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document.write( " </SPAN>\n" );
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