document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=397898>Question 397898</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do you solve for x in the following equation:\r<BR>\n" );
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document.write( "2^(2x+12) = 3^(x-45)\r<BR>\n" );
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document.write( "i changed it to log2 (2x+12) = log3 (x-45)<BR>\n" );
document.write( "is that right? where do i go from there? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #281900 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=281900\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2^(2x+12) = 3^(x-45)\r<BR>\n" );
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document.write( "i changed it to log2 (2x+12) = log3 (x-45)<BR>\n" );
document.write( "is that right? where do i go from there?<BR>\n" );
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document.write( "(2x+12)*log(2) = (x-45)*log(3) I think that's what you meant anyway.<BR>\n" );
document.write( "Just solve for x.  Logs of constants are constants.<BR>\n" );
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document.write( "2x*log(2) + 12log(2) = x*log(3) - 45log(3)<BR>\n" );
document.write( "2x*log(2) - x*log(3) = -12log(2) - 45log(3)<BR>\n" );
document.write( "x*(log(3) - 2log(2)) = 12log(2) + 45log(3)<BR>\n" );
document.write( "x = (12log(2) + 45log(3))/(log(3) - 2log(2))<BR>\n" );
document.write( "x =~ -200.7609\r<BR>\n" );
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document.write( " </SPAN>\n" );
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