document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=152746>Question 152746</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how would i solve this by hand? i know from the calcualator the answer is -3 but i'd like to learn how to do it the other way.\r<BR>\n" );
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document.write( "-2 ln e^2 + 1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #112289 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=112289\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> -2 ln e^2 + 1<BR>\n" );
document.write( "Write the log equiv of exponents: ln(e^2) = 2*ln(e)<BR>\n" );
document.write( "So you have:<BR>\n" );
document.write( "-2 * 2*ln(e) + 1<BR>\n" );
document.write( "Which is<BR>\n" );
document.write( "-4*ln(e) + 1<BR>\n" );
document.write( "The nat log of e = 1, therefore<BR>\n" );
document.write( "-4*1 + 1<BR>\n" );
document.write( "-4 + 1 = -3<BR>\n" );
document.write( " </SPAN>\n" );
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