document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=503492>Question 503492</A>: <I><SPAN STYLE=\"word-break: break-all;\"> i need help with 3 + log(2x+5)=2.... can you show me step by step and explain it to me please? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #339297 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=339297\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3 + log(2x+5)=2<BR>\n" );
document.write( "solving for x<BR>\n" );
document.write( "log(2x+5)=2-3=-1<BR>\n" );
document.write( "convert to exponential form: base(10) raised to log of number(-1)=number(2x+5)<BR>\n" );
document.write( "10^-1=2x+5<BR>\n" );
document.write( "1/10=2x+5<BR>\n" );
document.write( "20x+50=1<BR>\n" );
document.write( "20x=-49<BR>\n" );
document.write( "x=-49/20<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "3+log(2*49/20+5)<BR>\n" );
document.write( "3+(-1)=2 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );