document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=430877>Question 430877</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 5^(x-3)*3^(2x-8)=225<BR>\n" );
document.write( "X=? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #299141 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=299141\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 5^(x-3)*3^(2x-8)=225<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Take logs of both sides of equation and using the rules of exponents and multiplication.<BR>\n" );
document.write( "(x-3) log5+(2x-8) log3=log225<BR>\n" );
document.write( "xlog5-3log5+2xlog3-8log3=log225<BR>\n" );
document.write( "xlog5+2xlog3=log225+3log5+8log3<BR>\n" );
document.write( "x(log5+2log3)=log225+3log5+8log3<BR>\n" );
document.write( "x=(log225+3log5+8log3)/(log5+2log3)=5\r<BR>\n" );
document.write( "\n" );
document.write( "Check:<BR>\n" );
document.write( "5^2*3^2=25*9=225 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );