document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1077644>Question 1077644</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A new truck has a value of $36000. It depreciates in value 12% each year. To the nearest tenth of a year, when will the truck be worth 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 #692109 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=692109\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P=Po(1-0.12)^n<BR>\n" );
document.write( "Po=2*P<BR>\n" );
document.write( "Therefore, (0.5)=.88^n<BR>\n" );
document.write( "log of both sides<BR>\n" );
document.write( "-.3010=n log (.88)<BR>\n" );
document.write( "divide both sides by log .88<BR>\n" );
document.write( "n=5.4 years.<BR>\n" );
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