document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=968065>Question 968065</A>: <I><SPAN STYLE=\"word-break: break-all;\"> i need help with this question:\r<BR>\n" );
document.write( "\n" );
document.write( "The number of bacteria doubles after 3 hours. If there are N bacteria to start with, find the number of bacteria in 24 hours? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #591649 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=591649\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it doubles every 3 hours.<BR>\n" );
document.write( "this can be solved using either a continuous compounding function of f = p * e^(rn), or it can be solved using a discrete compounding formula of f = p * (1+r)^n.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "e is the scientific constant that is equal to 2.718281828.<BR>\n" );
document.write( "it's an irrational number that has an endless number of decimal digits.<BR>\n" );
document.write( "most often it is shown as such because the display number of digits on the calculator has a limitation of around 9 decimal digits.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the continuous compounding function is the one most often used, but i'll show you how to do it with both.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "using the continuous compounding formula, you would do the following.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the equation is f = p * e^(rn).<BR>\n" );
document.write( "the number of bacterial doubles in 3 hours, so you would set:<BR>\n" );
document.write( "p = 1<BR>\n" );
document.write( "f = 2<BR>\n" );
document.write( "n = 3<BR>\n" );
document.write( "the formula of f = p * e^(rn) would become:<BR>\n" );
document.write( "2 = 1 * e^(rn)<BR>\n" );
document.write( "you would now need to solve for r.<BR>\n" );
document.write( "you would normally divide both sides of the equation by p to get:<BR>\n" );
document.write( "f/p = e^(rn) would become:<BR>\n" );
document.write( "2 = e^(3r).<BR>\n" );
document.write( "you would then take the natural log of both sides of the equation to get:<BR>\n" );
document.write( "ln(2) = ln(e^(3r)).<BR>\n" );
document.write( "since ln(e^(3r)) is equivalent to 3*ln(e), and since ln(e) is equal to 1, the equation would then become:<BR>\n" );
document.write( "ln(2) = 3r.<BR>\n" );
document.write( "you would then divide by 3 to get:<BR>\n" );
document.write( "ln(2)/3 = r<BR>\n" );
document.write( "you would then solve for r to get:<BR>\n" );
document.write( "r = ln(2)/3 = .2310490602.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can confirm by replacing r in the original equation with .2310490602.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = p * e^(rn) would become 2 = 1 * e^(.2310490602 * 3) which would then become 2 = 2.<BR>\n" );
document.write( "this would confirm the solution for r is correct.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "now that you found r, you would use that to solve the problem.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "go back to the same equation and do the following:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = p * e^(rn)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "set n = 24<BR>\n" );
document.write( "set p = 1<BR>\n" );
document.write( "set r = .2310490602\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "equation becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = 1 * e^(.2310490602 * 24)<BR>\n" );
document.write( "solve for f to get:<BR>\n" );
document.write( "f = 256\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the bacteria will be 256 times the original amount in 24 hours.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the continuous compounding formula will be the one most often used, but you can solve this problem using the discrete compounding formula as well.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "here's how you would solve it using the discrete compounding formula.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that formula is:<BR>\n" );
document.write( "f = p * (1+r)^n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would do the following:<BR>\n" );
document.write( "f = 2<BR>\n" );
document.write( "p = 1<BR>\n" );
document.write( "n = 3<BR>\n" );
document.write( "the formula of f = p * (1+r)^n would become 2 = 1 * (1+r)^3<BR>\n" );
document.write( "you would divide both sides of the equation by p to get:<BR>\n" );
document.write( "2/1 = (1+r)^3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "note that, since p = 1, this step is not necessary, but p is not always equal to 1, so i'm showing you that step anyway.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "2/1 = (1+r)^3 would become:<BR>\n" );
document.write( "2 = (1+r)^3<BR>\n" );
document.write( "you would raise both sides of tghe equation to the 1/3 power to get:<BR>\n" );
document.write( "2^(1/3) = ((1+r)^3)^(1/3)<BR>\n" );
document.write( "since ((1+r)^3)^(1/3) is equal to (1+r)^(3*1/3) which is e3qual to (1+r)^1 which is equal to 1+r, your equation would become:<BR>\n" );
document.write( "2^(1/3) = 1 + r<BR>\n" );
document.write( "solve for 1+r to get:<BR>\n" );
document.write( "1+r = 2^(1/3) = 1.25992105<BR>\n" );
document.write( "subtract 1 from that to get:<BR>\n" );
document.write( "r = .25992105.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can confirm by replacing r in th eoriginal equation to get:<BR>\n" );
document.write( "2 = 1 * (1.25992105)^3 becomes 2 = 2.<BR>\n" );
document.write( "this confirms the value of r is good.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "now that you found r, you can use it to solve the problem.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "set p = 1 and set n = 24 and set r = .25992105.<BR>\n" );
document.write( "formula of f = 1 * (1+r)^n becomes:<BR>\n" );
document.write( "f = 1 * (1.25992105)^24<BR>\n" );
document.write( "solve for f to get:<BR>\n" );
document.write( "f = 256.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you got the same answer, as you should if you did the process correctly.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the continuous compounding formula is, once again.<BR>\n" );
document.write( "f = p * e^(rn)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "e is the scientific constant of 2.718281828<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the discrete compounding formuls is, once agakin.<BR>\n" );
document.write( "f = p * (1+r)^n\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "in your problem, the time periods were expressed in hours.<BR>\n" );
document.write( "n was therefore the number of hours.<BR>\n" );
document.write( "r was therefore the interest rate per hour.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );