document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=151233>Question 151233</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help. These log word problems are the worst.\r<BR>\n" );
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document.write( "A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #111130 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=111130\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A certain radioactive isotope decays at a rate of .25% annually. Determine the half-life of this isotope, to the nearest year.<BR>\n" );
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document.write( "The half-life formula: A = Ao*2^(-t/h)<BR>\n" );
document.write( "where:<BR>\n" );
document.write( "Ao - initial amt<BR>\n" );
document.write( "A = resulting 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( "In this problem:<BR>\n" );
document.write( "A = .75 (amt remaining after a 25% loss)<BR>\n" );
document.write( "Ao = 1<BR>\n" );
document.write( "t = 1<BR>\n" );
document.write( "h = half life <BR>\n" );
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document.write( "We can write it:<BR>\n" );
document.write( "1* 2(-1/h) = .75<BR>\n" );
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document.write( "ln(2^(-1/h)) = ln(.75); Find the nat log of both sides<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%2Fh%29 ALIGN=MIDDLE  ALT=\"%28-1%2Fh%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >*ln(2) = ln(.75); the log equiv of exponents<BR>\n" );
document.write( ".693147*<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2Fh ALIGN=MIDDLE  ALT=\"-1%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -.287682<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-.693147%2Fh ALIGN=MIDDLE  ALT=\"-.693147%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -.287682<BR>\n" );
document.write( "-.693147 = -.287682h; multiplied both sides by h<BR>\n" );
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document.write( "h = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-.693147%29%2F%28-.287682%29 ALIGN=MIDDLE  ALT=\"%28-.693147%29%2F%28-.287682%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = +2.4 years is the half-life of this isotope<BR>\n" );
document.write( "h = 2 yrs<BR>\n" );
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document.write( "Check solution on a calc: enter 2^(-1/2.4) = .749 ~ .75 (left after 25% loss) </SPAN>\n" );
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