document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1074574>Question 1074574</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A haddock has a length of 21 centimeters. Use Bertalanffy’s equation to predict, to the nearest tenth of a year, the age of the haddock. Assume m =94 centimeters, Lo =0.6 centimeter, and r =0.21.\r<BR>\n" );
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document.write( "L = m - (m - Lo)e^-rx  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #689381 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=689381\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> According to the question, we have:<BR>\n" );
document.write( "21=94-(94-0.6)e^-rx<BR>\n" );
document.write( "93.4e^-rx=73<BR>\n" );
document.write( "e^-rx=0.781585<BR>\n" );
document.write( "ln e^-rx=ln 0.781585<BR>\n" );
document.write( "-0.21x=-0.24643190408640580783666309806954<BR>\n" );
document.write( "x=1.173485257554313370650776657474 years as the age of the fish. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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