document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1189594>Question 1189594</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The mass of a substance, which follows a continuous exponential growth model, is being studied in a lab. The doubling time for this substance was observed to be 23 days. There were 72.6 mg of the substance present at the beginning of the study.\r<BR>\n" );
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document.write( "(a) Let t be the time (in days) since the beginning of the study, and let<BR>\n" );
document.write( "y be the amount of the substance at time t. Write a formula relating y to t. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.\r<BR>\n" );
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document.write( "(b) How much will be present in 8 days?<BR>\n" );
document.write( "Do not round any intermediate computations, and round your<BR>\n" );
document.write( "answer to the nearest tenth. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #821008 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=821008\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the formula is y = x * e ^ (r * t)<BR>\n" );
document.write( "to find the doubling time, let x = 1 and y = 2 and t = 23 days.<BR>\n" );
document.write( "the formula becomes:<BR>\n" );
document.write( "2 = 1 * e ^ (r * 23)<BR>\n" );
document.write( "take the natural log of both sides of the equation to get:<BR>\n" );
document.write( "ln(2) = ln(1 * e ^ (r * 23)<BR>\n" );
document.write( "since 1 * e = e, the formula becomes:<BR>\n" );
document.write( "ln(2) = ln(e ^ (r * 23)<BR>\n" );
document.write( "since ln(e ^ (r * 23) = r * 23 * ln(e) and since ln(3) = 1, the formula becomes:<BR>\n" );
document.write( "ln(2) = r * 23<BR>\n" );
document.write( "divide both sides of the equation by 23 to get:<BR>\n" );
document.write( "ln(2) / 23 = r<BR>\n" );
document.write( "solve for r to get:<BR>\n" );
document.write( "r = .0301368339.<BR>\n" );
document.write( "solve for y in the formula to confirm the dobuling time is correct.<BR>\n" );
document.write( "you get:<BR>\n" );
document.write( "y = 1 * e ^ (.0301368339 * 23) = 2.<BR>\n" );
document.write( "this confirms the doubling time is correct.<BR>\n" );
document.write( "when the starting value is 72.6, the doubling time will be:<BR>\n" );
document.write( "y = 72.6 * e ^ (.0301368339 * 23) = 145.2<BR>\n" );
document.write( "since 145.2 is double 72.6, the doubling time is correct when the initial value is 72.6.<BR>\n" );
document.write( "when the time from inception if 8 days, then the formula becomes:<BR>\n" );
document.write( "y = 72.6 * e ^ (.0301368339 * 8) = 92.39377381.<BR>\n" );
document.write( "round this to the nearest 10th to get 92.4<BR>\n" );
document.write( "that's your solution.<BR>\n" );
document.write( "the formula can be graphed by letting x = t.<BR>\n" );
document.write( "here's what the graph looks like.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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