document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1131>Question 1131</A>: <I><SPAN STYLE=\"word-break: break-all;\"> HOw do I solve  log16 2=w </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #361 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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=361\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> By definition of logarithm,log16 2= w means 16^w = 2,<BR>\n" );
document.write( " since 16 = 2^4, so (2^4)^w = 2.<BR>\n" );
document.write( " By exponential law ,(2^4)^w = 2^(4w) = 2 = 2^1.<BR>\n" );
document.write( " Comparing the exponent, we have 4w = 1, w = 1/4.\r<BR>\n" );
document.write( "\n" );
document.write( " Another way,use the rule of changing base (to base c)<BR>\n" );
document.write( " loga b = logc b/logc a.<BR>\n" );
document.write( " <BR>\n" );
document.write( " w = log16 2 = log2 2/ log2 16 (changing base from 16 to 2)<BR>\n" );
document.write( "         = 1 / log2 2^4<BR>\n" );
document.write( "         = 1 / (4 log2 2)<BR>\n" );
document.write( "         = 1/4 (same answer)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " I prefer to use the 2nd method. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );