document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=38872>Question 38872</A>: <I><SPAN STYLE=\"word-break: break-all;\"> i know its already solved but i need to know how to prove it\r<BR>\n" );
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document.write( "6log base 6^(3x-1)= (3x-1)\r<BR>\n" );
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document.write( "thanks a bunch! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #26148 by </B><A HREF=/tutors/aboutme.mpl?userid=DavidChandler><B>DavidChandler(2)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=DavidChandler><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=26148\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I think the problem you are trying to state is:\r<BR>\n" );
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document.write( "6^[log base 6 of (3x-1)]= (3x-1)\r<BR>\n" );
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document.write( "The exponential function and log function (for any base) are inverse functions of each other.  Therefore applying one then the other brings you back where you started.  In this case you start with te expression (3x-1), take its base 6 log, then plug the result into the base 6 exponential function.  The two steps cancel each other out.  It's like adding 2 then subtracting 2, or doubling then halving.  Inverse functions \"undo\" each other. </SPAN>\n" );
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