document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1101259>Question 1101259</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A person bought a cat for $21,028. After 4 years, the car was worth $11,987. If the value of the car continues to decrease exponentially at this rate then determine the time it will take to be worth $4992 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #715848 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=715848\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If the value decreases exponentially, then:<BR>\n" );
document.write( "11987=21028e^-4k; where k is the depreciation constant. So<BR>\n" );
document.write( ".57005=e^-4k<BR>\n" );
document.write( "ln .57005=5ln e^-4k=-4k ln e=-4k<BR>\n" );
document.write( "k=0.140508<BR>\n" );
document.write( "Then:<BR>\n" );
document.write( "4992/21028=e^-0.140508t<BR>\n" );
document.write( "ln 0.2374=ln e^-0.140508t=-0.140508t ln e=-0.140508t<BR>\n" );
document.write( "t=10.234 years for the car to be worth $4992<BR>\n" );
document.write( "&#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );