document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=332189>Question 332189</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If 525 mg of a radioactive element decays to 400 mg in 48 hours, what is its half-life? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #238121 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=238121\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If 525 mg of a radioactive element decays to 400 mg in 48 hours,<BR>\n" );
document.write( "what is its half-life?<BR>\n" );
document.write( ":<BR>\n" );
document.write( "The half-life formula:  Ao*2^(-t/h) = A<BR>\n" );
document.write( "Where<BR>\n" );
document.write( "A = amt after t (400)<BR>\n" );
document.write( "Ao = initial amt (525)<BR>\n" );
document.write( "t = time (48 hrs)<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
document.write( ":<BR>\n" );
document.write( "525*2^(-48/h) = 400<BR>\n" );
document.write( "Divide both sides by 525<BR>\n" );
document.write( "2^(-48/h) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=400%2F525 ALIGN=MIDDLE  ALT=\"400%2F525\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "2^(-48/h) = .762<BR>\n" );
document.write( "Use nat logs<BR>\n" );
document.write( "ln(2^(-48/h)) = ln(.762)<BR>\n" );
document.write( "log equiv of exponents<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-48%2Fh ALIGN=MIDDLE  ALT=\"-48%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" >*ln(2) = ln(.762)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-48%2Fh ALIGN=MIDDLE  ALT=\"-48%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" >= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28.762%29%2Fln%282%29 ALIGN=MIDDLE  ALT=\"ln%28.762%29%2Fln%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "using a calc<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-48%2Fh ALIGN=MIDDLE  ALT=\"-48%2Fh\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -.392<BR>\n" );
document.write( "-48 = -.392h<BR>\n" );
document.write( "h = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-48%29%2F%28-.392%29 ALIGN=MIDDLE  ALT=\"%28-48%29%2F%28-.392%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "h = +122.45 hr is the half-life of the substance<BR>\n" );
document.write( ":<BR>\n" );
document.write( ":<BR>\n" );
document.write( "Check this in the half-life formula, using a calc<BR>\n" );
document.write( "Ao*2^(-t/h) <BR>\n" );
document.write( "enter 525*2^(-48/122.45) results: 400.1 ~ 400; confirms our solution </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );