document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160470>Question 1160470</A>: <I><SPAN STYLE=\"word-break: break-all;\"> log 8 ≈ 0.9<BR>\n" );
document.write( "log 7 ≈ 0.8<BR>\n" );
document.write( "log 12 ≈ 1.1<BR>\n" );
document.write( "Find log 96<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #783733 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=783733\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "log(96) = log(12*8)<BR>\n" );
document.write( "log(96) = log(12)+log(8) ... use log rule log(x*y) = log(x)+log(y)<BR>\n" );
document.write( "log(96) = 1.1+0.9 approximately<BR>\n" );
document.write( "log(96) = <font color=red>2.0 approximately</font>\r<BR>\n" );
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document.write( "Using a calculator, the more accurate result is roughly log(96) = 1.98227123303957<BR>\n" );
document.write( "The error comes from the fact we rounded these two values <BR>\n" );
document.write( "log(8) = 0.90308998699194<BR>\n" );
document.write( "log(12) = 1.07918124604762<BR>\n" );
document.write( "to one decimal place<BR>\n" );
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