document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=237215>Question 237215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is a story problem I am stuck on could use some help.\r<BR>\n" );
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document.write( "A computer is infected with the sasser virus. Assume that it infects 20 other computers in 5 minutes; and that these PCs and servers infect 20 more machines within another five minutes, etc. How long until 100 million computers are infected? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #174494 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=174494\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 20 computers are infected every 5 minutes.\r<BR>\n" );
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document.write( "how long to infect 100 million computers.\r<BR>\n" );
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document.write( "this looks like a geometric progression with the common ratio of r = 20\r<BR>\n" );
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document.write( "the last term in a geometric progression is given by the formula:\r<BR>\n" );
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document.write( "x[n] = x[0] * r^(n-1)\r<BR>\n" );
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document.write( "We know x[0] and we know x[n] and we know r.\r<BR>\n" );
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document.write( "we need to solve for n.\r<BR>\n" );
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document.write( "Our formula becomes:\r<BR>\n" );
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document.write( "100 million = 1 * 20^(n-1)\r<BR>\n" );
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document.write( "because:\r<BR>\n" );
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document.write( "x[0] = 1<BR>\n" );
document.write( "r = 20<BR>\n" );
document.write( "x[n] = 100 million\r<BR>\n" );
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document.write( "this equation becomes:\r<BR>\n" );
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document.write( "100 million = 20^(n-1)\r<BR>\n" );
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document.write( "we take the log of both sides of this equation to get:\r<BR>\n" );
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document.write( "log(100 million) = log(20^(n-1))\r<BR>\n" );
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document.write( "by the laws of logarithms, this becomes:\r<BR>\n" );
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document.write( "log(100 million) = (n-1)*log(20)\r<BR>\n" );
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document.write( "divide both sides of this equation by log(20) to get:\r<BR>\n" );
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document.write( "log(100 million) / log(20) = n-1\r<BR>\n" );
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document.write( "solve for (n-1) to get:\r<BR>\n" );
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document.write( "n-1 = 6.148974295\r<BR>\n" );
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document.write( "this means that 100 million computers will be infected in (n-1) * 5 minutes which equals 6.148974295 * 5 = 30.74487147\r<BR>\n" );
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document.write( "to see how this works, do the calculations for each n.\r<BR>\n" );
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document.write( "0 minutes = 1 * 20^0 = 1<BR>\n" );
document.write( "5 minutes = 1 * 20^1 = 20<BR>\n" );
document.write( "10 minutes = 1 * 20^2 = 400<BR>\n" );
document.write( "15 minutes = 1 * 20^3 = 8000<BR>\n" );
document.write( "20 minutes = 1 * 20^4 = 160 thousand<BR>\n" );
document.write( "25 minutes = 1 * 20^5 = 3.2 million<BR>\n" );
document.write( "30 minutes = 1 * 20^6 = 64 million<BR>\n" );
document.write( "30.74487147 minutes = 1 * 20^6.148974295 = 100 million\r<BR>\n" );
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document.write( "this can also be solved using the growth formula of:\r<BR>\n" );
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document.write( "future = base * e^kt\r<BR>\n" );
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document.write( "where e = the scientific constant of 2.718281828....., and k = the constant of growth.\r<BR>\n" );
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document.write( "since future = 100 million and base = 1, this formula would become:\r<BR>\n" );
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document.write( "100 million = 1 * e^kt\r<BR>\n" );
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document.write( "we would need, however, to establish what k is.\r<BR>\n" );
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document.write( "since we know that when t = 5, future = 20, our formula would become:\r<BR>\n" );
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document.write( "20 = 1*e^5k where:\r<BR>\n" );
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document.write( "future = 20<BR>\n" );
document.write( "base = 1<BR>\n" );
document.write( "k = unknown<BR>\n" );
document.write( "t = 5 minutes\r<BR>\n" );
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document.write( "we solve this equation for k to get:\r<BR>\n" );
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document.write( "20 = e^5k<BR>\n" );
document.write( "take natural log of both sides to get:<BR>\n" );
document.write( "ln(20) = ln(e^5k)<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "ln(20) = ln(e) * 5k<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "ln(20) = 5k<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "k = ln(20)/5<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "k = .599146455\r<BR>\n" );
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document.write( "our original equation becomes:\r<BR>\n" );
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document.write( "100 million = e^(.599146455*t)\r<BR>\n" );
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document.write( "take ln of both sides of the equation to get:\r<BR>\n" );
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document.write( "ln(100 million) = ln(e^(.599146455*t)<BR>\n" );
document.write( "this becomes:<BR>\n" );
document.write( "ln(100 million)/.599146455 = t<BR>\n" );
document.write( "we solve for t to get:<BR>\n" );
document.write( "t = 30.74487147\r<BR>\n" );
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document.write( "since this is the same answer we got using the geometric progression, we did good.\r<BR>\n" );
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document.write( "the use of the constant \"e\" simplifies the operation somewhat if you understand that you have to solve for the constant k first, but you can solve it either way and get the same answer.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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