document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=33366>Question 33366</A>: <I><SPAN STYLE=\"word-break: break-all;\"> log(10x/y)=1+log x-log y.  I have no idea whether this problem is true or false.  Could you please help show me the steps in finding the answer? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #19823 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=19823\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You need to know the Laws of logarithms.<BR>\n" );
document.write( "log(10x/y)= log(10x)-logy<BR>\n" );
document.write( "=log10+logx-logy<BR>\n" );
document.write( "=1+logx-logy<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );