document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=767375>Question 767375</A>: <I><SPAN STYLE=\"word-break: break-all;\"> i''m having trouble understanding how these problems work this is the problem.. S = 12.75 e^(-0.0425 t). When S=7.35, find t. i dont know were to start can you please help me </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #467582 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><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=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=467582\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> S = 12.75 e^-0.0425t<BR>\n" );
document.write( "Find t when S = 7.35<BR>\n" );
document.write( "S = 12.75 e^-0.0425t<BR>\n" );
document.write( "7.35 = 12.75 e^-0.0425t<BR>\n" );
document.write( "7.35/12.75 = e^-0.0425t<BR>\n" );
document.write( "0.57647 = -0.0425t e<BR>\n" );
document.write( "log e both sides<BR>\n" );
document.write( "(Notice how the -0.0425t comes in front of e)<BR>\n" );
document.write( "loge 0.57647 = -0.0425t loge e<BR>\n" );
document.write( "(loge e = 1)<BR>\n" );
document.write( "loge 0.57647/-0.0425 = t<BR>\n" );
document.write( "t = 12.96<BR>\n" );
document.write( "Hope this helps.<BR>\n" );
document.write( ":-) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );