document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=104685>Question 104685</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Could someone please assist me by solving this logarithmic equation step-by-step for future reference. Thank you.\r<BR>\n" );
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document.write( "log(7y + 1) = 2log(y+3)-log2\r<BR>\n" );
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document.write( "log(7y + 1) = log(y+3)^2-log2\r<BR>\n" );
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document.write( "log(7y + 1) = log(y+3)^2/2<BR>\n" );
document.write( "you can write<BR>\n" );
document.write( "(7y + 1) = (y+3)^2/2<BR>\n" );
document.write( "2(7y + 1) = y^2+6y+9<BR>\n" );
document.write( "14y+2=y^2+6y+9<BR>\n" );
document.write( "y^2-8y+7=0\r<BR>\n" );
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document.write( "y1=(8+rootsquare(36))/2<BR>\n" );
document.write( "  =(8+6)/2<BR>\n" );
document.write( "  =7<BR>\n" );
document.write( "y2=(8-rootsquare(36))/2<BR>\n" );
document.write( "  =(8-6)/2<BR>\n" );
document.write( "  =1\r<BR>\n" );
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document.write( "so (y1,y2) are (7,1) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #76234 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=76234\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> handy rules for logs:<BR>\n" );
document.write( "adding logs is multiplying numbers<BR>\n" );
document.write( "subtracting logs is dividing numbers<BR>\n" );
document.write( "multiplying logs is exponentiation<BR>\n" );
document.write( "dividing logs is taking roots\r<BR>\n" );
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document.write( "with those in mind ... 7y+1=((y+3)^2)/2 ... 14y+2=y^2+6y+9\r<BR>\n" );
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document.write( "y^2-8y+7=0 ... (y-7)(y-1)=0 </SPAN>\n" );
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