document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=581589>Question 581589</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do I solve the following using the relationship between natural logarithms, Ins, and the number e?<BR>\n" );
document.write( "1)In(2m+3)=8<BR>\n" );
document.write( "2)1.1+In x^2=6<BR>\n" );
document.write( "3)e^x+1=30 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #371714 by </B><A HREF=/tutors/aboutme.mpl?userid=nerdybill><B>nerdybill(7384)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nerdybill><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=nerdybill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=371714\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)In(2m+3)=8<BR>\n" );
document.write( "2m+3 = e^8<BR>\n" );
document.write( "2m = e^8 - 3<BR>\n" );
document.write( "m = (e^8 - 3)/2<BR>\n" );
document.write( "m = 1488.98<BR>\n" );
document.write( ".<BR>\n" );
document.write( "2)1.1+In x^2=6<BR>\n" );
document.write( "Ln x^2 = 6 - 1.1<BR>\n" );
document.write( "x^2 = e^(6 - 1.1)<BR>\n" );
document.write( "x = sqrt(e^(6 - 1.1))<BR>\n" );
document.write( "x = 11.59<BR>\n" );
document.write( ".<BR>\n" );
document.write( "3)e^x+1=30<BR>\n" );
document.write( "e^x=30-1<BR>\n" );
document.write( "x = ln(30-1)<BR>\n" );
document.write( "x = 3.37 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );