document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=821092>Question 821092</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do i solve A=ln2 in terms of A<BR>\n" );
document.write( "ln(16)<BR>\n" );
document.write( "ln(square root of 2)<BR>\n" );
document.write( "ln(1/2) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #493898 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=493898\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> How do i solve A=ln2 in terms of A<BR>\n" );
document.write( "2 = e^A<BR>\n" );
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document.write( "ln(16) = ln[(2^4)] = ln[(e^A)^4] = ln[e^(4A)] = 4A<BR>\n" );
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document.write( "ln(square root of 2) = ln[2^(1/2)] = (1/2)ln(2) = (1/2)ln[e^A] = (1/2)A<BR>\n" );
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document.write( "ln(1/2) = ln[2^-1] = ln[(e^A)^-1] = ln[e^(-A)] = -A<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "==================== </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );