document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=362852>Question 362852</A>: <I><SPAN STYLE=\"word-break: break-all;\"> hi how would i go about solving 9^(x-5)=81^(x-7) for logarithms i know that the 81 would turn to 9 to have a common base but im not sure what else to do with the powers? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #258655 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=258655\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 9^(x-5)=81^(x-7)<BR>\n" );
document.write( "9^(x-5)=9^2(x-7)<BR>\n" );
document.write( "since the bases are same<BR>\n" );
document.write( "x-5 = 2(x-7)<BR>\n" );
document.write( "x-5=2x-14<BR>\n" );
document.write( "2x-x=14-5<BR>\n" );
document.write( "x=9<BR>\n" );
document.write( "...<BR>\n" );
document.write( "m.ananth@hotmail.ca </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );