document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1091248>Question 1091248</A>: <I><SPAN STYLE=\"word-break: break-all;\"> When the meteor hit the earth, there were 600,000 dinosaurs alive. Thirty years later there were only 15,000. If they died exponentially, after how many years would there be only five dinosaurs left?\r<BR>\n" );
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document.write( "---I believe the exponential growth/decay formula I learned is y=k&#8901;a^t, in this case I think it would be y=600,000(1-15,000)^30  \r<BR>\n" );
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document.write( "y = amount after increase/decrease (in this case decrease)\r<BR>\n" );
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document.write( "k = original amount\r<BR>\n" );
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document.write( "a = decrease/decay factor. In this case I think it would be (1-15,000)\r<BR>\n" );
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document.write( "t = time \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #705639 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=705639\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 600000*b^t, where t=30=15000<BR>\n" );
document.write( "b^30=15000/600000=1/40<BR>\n" );
document.write( "30 ln b=ln (1/40)=-3.689<BR>\n" );
document.write( "divide by 30 without rounding<BR>\n" );
document.write( "ln b=-0.1230.<BR>\n" );
document.write( "raise both sides to the e power<BR>\n" );
document.write( "b=0.8843<BR>\n" );
document.write( "600000*0.8843^t=5<BR>\n" );
document.write( "divide by 600000<BR>\n" );
document.write( "0.8843^x=5/600000=0.000008333<BR>\n" );
document.write( "logs of both sides<BR>\n" );
document.write( "x log (0.8843)=-5.079<BR>\n" );
document.write( "x=-5.079/log (0.8843)=95.12 years ANSWER<BR>\n" );
document.write( "check<BR>\n" );
document.write( "in 30 years, it decreases 1/40<BR>\n" );
document.write( "in 60 years, 1/1600<BR>\n" );
document.write( "in 90 years, 1/64000<BR>\n" );
document.write( "so it will decrease somewhat more than 1/64000<BR>\n" );
document.write( "600000*(0.8843)^95=5.07 </SPAN>\n" );
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