document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=327293>Question 327293</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A certian population obeys the latural law of decay. at a certian time,the population is 40,000. three years later the population is 38000. approximately how long does it take the population to decline from 40,000 to 20,000? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #234421 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=234421\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A certain population obeys the natural law of decay. at a certian time,the population is 40,000. three years later the population is 38000. approximately how long does it take the population to decline from 40,000 to 20,000?<BR>\n" );
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document.write( "You have two points (0,40000) and (3,38000)<BR>\n" );
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document.write( "Assuming the decay is exponential, use the <BR>\n" );
document.write( "model A(t) = ab^t<BR>\n" );
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document.write( "Solve for a and b<BR>\n" );
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document.write( "Using (0,40000), solve for \"a\":<BR>\n" );
document.write( "40,000 = ab^0<BR>\n" );
document.write( "a = 40,000<BR>\n" );
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document.write( "Now, A(t) = 40,000*b^t<BR>\n" );
document.write( "Using (3,38000), solve for \"b\":<BR>\n" );
document.write( "38000, = 40,000*b^3<BR>\n" );
document.write( "19/20 = b^3<BR>\n" );
document.write( "b = 0.983<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "A(t) = 40000*0.983^t<BR>\n" );
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document.write( "When does the population decay to 20,000?<BR>\n" );
document.write( "20,000 = 40000*0.983^t<BR>\n" );
document.write( "0.983^t = 0.5<BR>\n" );
document.write( "t = log(0.5)/log(0.983)<BR>\n" );
document.write( "t is approximately 40<BR>\n" );
document.write( "Rounded up you would get 41 years<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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