document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=913852>Question 913852</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you show that (8^n+3^n)/(8^-n+3^-n)=24^n without using the logarith way? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #554765 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=554765\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (8^n + 3^n) / (8^-n + 3^-n)<BR>\n" );
document.write( "work with the denominator<BR>\n" );
document.write( "(8^-n + 3^-n) = (1/8^n + 3^n)<BR>\n" );
document.write( "LCM is 24, therefore<BR>\n" );
document.write( "(8^-n + 3^-n) = (3^n + 8^n) / 24^n<BR>\n" );
document.write( "invert denominator and multiply by numerator<BR>\n" );
document.write( "(8^n + 3^n) * 24^n / (8^n + 3^n) = 24^n<BR>\n" );
document.write( "note that denominator and numerator cancels (8^n + 3^n)<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );