document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=590236>Question 590236</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.	Show that log8x=1/3 log2x hence solve the equation for x>0<BR>\n" );
document.write( "               Log2(3x+1) +log8(x-1)3=0<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #375083 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=375083\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1.	Show that log8x=1/3 log2x hence solve the equation for x>0<BR>\n" );
document.write( "Log2(3x+1) +log8(x-1)3=0<BR>\n" );
document.write( "**<BR>\n" );
document.write( "log8x=1/3 log2x<BR>\n" );
document.write( "change to base2<BR>\n" );
document.write( "log8x=log2x/log2(8)=log2x/3<BR>\n" );
document.write( "..<BR>\n" );
document.write( "log2(3x+1) +log8(x-1)*3=0<BR>\n" );
document.write( "change to base 2<BR>\n" );
document.write( "log2(3x+1) +log2(x-1)*3/log2(8)=0<BR>\n" );
document.write( "log2(3x+1) +log2(x-1)*3/3=0<BR>\n" );
document.write( "log2(3x+1) +log2(x-1)=0<BR>\n" );
document.write( "place under single log<BR>\n" );
document.write( "log2[(3x+1)(x-1)]=0<BR>\n" );
document.write( "convert to exponential form:<BR>\n" );
document.write( "2^0=(3x+1)(x-1)=1<BR>\n" );
document.write( "3x^2-2x-2=0<BR>\n" );
document.write( "solve by quadratic formula:<BR>\n" );
document.write( "x&#8776;-.549 (reject,x>0)<BR>\n" );
document.write( "or<BR>\n" );
document.write( "x=1.215\r<BR>\n" );
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