document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=984029>Question 984029</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A machine depreciates in value each year by 8% of its value at the beginning of the year. After how many years will its value first be less than half its original value? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #604826 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=604826\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is basically an interest problem with the percentage negative.<BR>\n" );
document.write( "P=Po{1-0.08}^n<BR>\n" );
document.write( "But P/P0=0.5, since it will be half as much.<BR>\n" );
document.write( "0.5=1.08^n<BR>\n" );
document.write( "take ln of both sides<BR>\n" );
document.write( "0.693=n ln (1.08); ln (1.08)=0.07696 (rounded).<BR>\n" );
document.write( "Not rounding, <BR>\n" );
document.write( "divide by that.<BR>\n" );
document.write( "9.006 years (rounded)<BR>\n" );
document.write( "This is the rule of 72 in reverse.   </SPAN>\n" );
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