document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=487640>Question 487640</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I don't actually understand how to do logarithms, I've been taught but I cannot understand them, so would you please be able to help me by showing me how to solve for x: log x + log 2x = log 50.\r<BR>\n" );
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document.write( "Thanking You.<BR>\n" );
document.write( "Tracey </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #333094 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=333094\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log x + log 2x = log 50<BR>\n" );
document.write( "log x + log 2x - log 50=0<BR>\n" );
document.write( "place under a single log using log rules of multiplication and division<BR>\n" );
document.write( "log(x*2x/50)=0<BR>\n" );
document.write( "Convert to exponential form: base(10) raised to log of number(0)=number (x*2x/50)<BR>\n" );
document.write( "10^0=(x*2x/50)<BR>\n" );
document.write( "1=2x^2/50<BR>\n" );
document.write( "2x^2=50<BR>\n" );
document.write( "x^2=50/2=25<BR>\n" );
document.write( "x=±&#8730;25=±5<BR>\n" );
document.write( "x=-5 (reject, x>0)<BR>\n" );
document.write( "or<BR>\n" );
document.write( "x=5 (ans)<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "log x + log 2x = log 50<BR>\n" );
document.write( "log 5+log10=log 5*10=log 50 </SPAN>\n" );
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