document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=971492>Question 971492</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the population of a town is 32,000 in the year 2013, with a growing rate of 1.2 percent per year, when will the population reach 40,000 & what will the population be in 2020 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #802411 by </B><A HREF=/tutors/aboutme.mpl?userid=CubeyThePenguin><B>CubeyThePenguin(3113)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CubeyThePenguin><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=CubeyThePenguin><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=802411\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the number of years that have passed.\r<BR>\n" );
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document.write( "32,000(1.012)^x = 40,000\r<BR>\n" );
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document.write( "x = 18.7066\r<BR>\n" );
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document.write( "The population reaches 40,000 in 2031.\r<BR>\n" );
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document.write( "In 2020, x = 7.\r<BR>\n" );
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document.write( "32,000(1.012)^7 = 34,786 </SPAN>\n" );
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