document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=936075>Question 936075</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do I solve: it takes 12 hrs for a certain bacterial culture to double in size. How long will it take the same bacterial culture to triple? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #569513 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=569513\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula is f = p * e^kt\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "k is the constant of growth or decay<BR>\n" );
document.write( "if k is positive, it is the constant of growth.<BR>\n" );
document.write( "if k is negative, it is the constant of decay.<BR>\n" );
document.write( "t is equal to the time interval.\r<BR>\n" );
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document.write( "for it to double, you have the following:<BR>\n" );
document.write( "f = 2<BR>\n" );
document.write( "p = 1<BR>\n" );
document.write( "t = 12 hours\r<BR>\n" );
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document.write( "you need to use this equation to solve for k.\r<BR>\n" );
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document.write( "formula becomes:\r<BR>\n" );
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document.write( "2 = 1 * e^(k*12)\r<BR>\n" );
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document.write( "start with:\r<BR>\n" );
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document.write( "2 = e^(k*12)\r<BR>\n" );
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document.write( "take the natural log of both sides of the equation to get:\r<BR>\n" );
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document.write( "ln(2) = ln(e^(k*12))\r<BR>\n" );
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document.write( "since ln(e^(k*12)) = k*12*ln(e), and since ln(e) = 1, you get:\r<BR>\n" );
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document.write( "ln(2) = k * 12\r<BR>\n" );
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document.write( "divide both sides of this equation by 12 to get:\r<BR>\n" );
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document.write( "ln(2) / 12 = k\r<BR>\n" );
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document.write( "solve for k to get:\r<BR>\n" );
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document.write( "k = .057762265\r<BR>\n" );
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document.write( "confirm by replacing k in the original equation to confirm that the equation is true.\r<BR>\n" );
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document.write( "you get:\r<BR>\n" );
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document.write( "2 = 1 * e^(.057762265*12) which becomes 2 = 2.\r<BR>\n" );
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document.write( "this confirms the solution for k is good.\r<BR>\n" );
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document.write( "now you can use k to solve for how long it will take the same bacterial culture to triple.\r<BR>\n" );
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document.write( "equation now becomes:\r<BR>\n" );
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document.write( "3 = 1 * e^(.057762265*t)\r<BR>\n" );
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document.write( "you are solving for t.\r<BR>\n" );
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document.write( "start with:\r<BR>\n" );
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document.write( "3 = e^(.057762265*t)\r<BR>\n" );
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document.write( "take the natural log of both sides of the equation to get:\r<BR>\n" );
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document.write( "ln(3) = ln(e^(.057762265*t) which becomes:\r<BR>\n" );
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document.write( "ln(3) = .057762265*t*ln(e) which becomes:\r<BR>\n" );
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document.write( "ln(3) = .057762265*t\r<BR>\n" );
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document.write( "divide both sides of the equation by .057762265 to get:\r<BR>\n" );
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document.write( "ln(3) / .057762265 = t\r<BR>\n" );
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document.write( "solve for t to get:\r<BR>\n" );
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document.write( "t = 19.01955001\r<BR>\n" );
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document.write( "confirm by replacing t in the original equation to see if that equation is true.\r<BR>\n" );
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document.write( "original equation is 3 = 1 * e^(.057762265 * 19.01955001) which becomes 3 = 3.\r<BR>\n" );
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document.write( "this confirms the solution is correct.\r<BR>\n" );
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document.write( "your solution is that it will take 19.01955001 hours for the solution to triple.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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