document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=562263>Question 562263</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm having trouble solving this problem log 2x =-1. Can you give me an example of a problem close to this one? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #364510 by </B><A HREF=/tutors/aboutme.mpl?userid=issacodegard><B>issacodegard(60)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=issacodegard><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=issacodegard><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=364510\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi, let say that log_b is the logarithm with base b function.<BR>\n" );
document.write( "Solve log_2(3y+1)=6 for y<BR>\n" );
document.write( "We have,<BR>\n" );
document.write( "log_2(3y+1)=6<BR>\n" );
document.write( "2^(log_2(3y+1))=2^(6)  (exponentiate with respect to the base 2)<BR>\n" );
document.write( "3y+1=64<BR>\n" );
document.write( "3y=63<BR>\n" );
document.write( "y=21\r<BR>\n" );
document.write( "\n" );
document.write( "Note that log_b(x) and b^x are inverse functions, so b^(log_b(x))=x </SPAN>\n" );
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