document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1082600>Question 1082600</A>: <I><SPAN STYLE=\"word-break: break-all;\"> `Can someone please help me solve this problem: In 2000 the population of a town was 35000. In 2010 it was 50000. Find an exponential model of the form f(t)=A base 0 e^kt for this population, where t is time in years since 2000. I also need help finding K but I think you have to use Ln logarithm but I am not sure. \r<BR>\n" );
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document.write( "B) what year would the population reach 100,000?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #696647 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=696647\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The exponential growth/decay model is<BR>\n" );
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document.write( "A = A(o) * e^kt, where A(o) is the amount we start with, k is the constant of growth or decay, t is time <BR>\n" );
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document.write( "A) 50000 = 35000 * e^10k<BR>\n" );
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document.write( "e^10k = 50000 / 35000 = 1.4286<BR>\n" );
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document.write( "10k = ln 1.4286 = 0.3567<BR>\n" );
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document.write( "k = 0.3567 / 10 = 0.03567<BR>\n" );
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document.write( "B)  100000 = 35000 * e^(0.03567t)<BR>\n" );
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document.write( "e^(0.03567t) = 100000 / 35000 = 2.8571<BR>\n" );
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document.write( "0.03567t = ln 2.8571 = 1.0498<BR>\n" );
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document.write( "t = 1.0498 / 0.03567 = 29.4309 approx 29<BR>\n" );
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document.write( "the population will be 100000 in year 2029<BR>\n" );
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