document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=614548>Question 614548</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for x:<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+2%283%5Ex%29+=+7%285%5Ex%29+ ALIGN=MIDDLE  ALT=\"+2%283%5Ex%29+=+7%285%5Ex%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #386601 by </B><A HREF=/tutors/aboutme.mpl?userid=sophxmai><B>sophxmai(62)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=sophxmai><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=sophxmai><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=386601\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> take the log of both sides and then separate the 2(3^x) by the log(m)+log(n)=log(mn) rule.\r<BR>\n" );
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document.write( "log(2(3^x))=log(7(5^x))<BR>\n" );
document.write( "log(2)+log(3^x)=log(7)+log(5^x)<BR>\n" );
document.write( "log(2)+xlog(3)=log(7)+xlog(5)<BR>\n" );
document.write( "xlog(3)-xlog(5)=log(7)-log(2)<BR>\n" );
document.write( "x(log(3)-log(5))=log(7)-log(2)<BR>\n" );
document.write( "x=(log(7)-log(2))/(log(3)-log(5))<BR>\n" );
document.write( "x=-2.452427 <--i used a calculator\r<BR>\n" );
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document.write( "x=(log(7)-log(2))/(log(3)-log(5)) can be rewritten using the rule log(m)-log(n)=log(m/n)<BR>\n" );
document.write( "x=log(7/2)/log(3/5) </SPAN>\n" );
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