document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1060125>Question 1060125</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the exponential decay equation, where A is the amount of a radioactive material present after time t, k is the half-life of the radioactive material, and A(base 0) is the original amount of radioactive substance.\r<BR>\n" );
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document.write( "A 5 microgram sample of a radioactive isotope decays to 4.23 micrograms in 9 min. What is the half-life of the radioactive isotope, in minutes? (Round your answer to two decimal places.)<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #675329 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=675329\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the exponential decay equation, where A is the amount of a radioactive material present after time t, k is the half-life of the radioactive material, and A(base 0) is the original amount of radioactive substance.<BR>\n" );
document.write( "A = Ao*2^(-t/k)<BR>\n" );
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document.write( "A 5 microgram sample of a radioactive isotope decays to 4.23 micrograms in 9 min.<BR>\n" );
document.write( " What is the half-life of the radioactive isotope, in minutes?<BR>\n" );
document.write( "A = 5 micrograms<BR>\n" );
document.write( "Ao = 4.23<BR>\n" );
document.write( "t = 9 min<BR>\n" );
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document.write( "5 * 2(-9/k) = 4.23<BR>\n" );
document.write( "2(-9/k) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4.23%2F5 ALIGN=MIDDLE  ALT=\"4.23%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "using nat logs<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-9%2Fk ALIGN=MIDDLE  ALT=\"-9%2Fk\"   DISABLED_style= \"max-width:90%; height:auto;\" >ln(2) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%284.23%2F5%29 ALIGN=MIDDLE  ALT=\"ln%284.23%2F5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-9%2Fk ALIGN=MIDDLE  ALT=\"-9%2Fk\"   DISABLED_style= \"max-width:90%; height:auto;\" > * .693 = -.167<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-9%2Fk ALIGN=MIDDLE  ALT=\"-9%2Fk\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-.167%2F.693 ALIGN=MIDDLE  ALT=\"-.167%2F.693\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-9%2Fk ALIGN=MIDDLE  ALT=\"-9%2Fk\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -.241<BR>\n" );
document.write( "k = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-9%29%2F%28-.241%29 ALIGN=MIDDLE  ALT=\"%28-9%29%2F%28-.241%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " k = +37.34 days<BR>\n" );
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document.write( "Check this on your calc: enter 5 * 2^(-9/37.34) =\r<BR>\n" );
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document.write( " (Round your answer to two decimal places. </SPAN>\n" );
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