document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1202599>Question 1202599</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write an exponential model for the following situation. The drug dosage is 375 mg. The drug is eliminated at a rate of 11.3% per hour. Use D=the amount of the drug in milligrams and t=time in hours. Enter your model in the simplified form y=asup(bracket(b),x), and be mindful about the case of your variables.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #837527 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=837527\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the drug dose is 375 milligrams.<BR>\n" );
document.write( "the drug is eliminated at a rate of 11.3% per hour.\<BR>\n" );
document.write( "D = amount of drug in milligrams.<BR>\n" );
document.write( "t = time in hours.\r<BR>\n" );
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document.write( "the equation i use most for these if f = p * (1 + r) ^ n<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the growth rate per time period<BR>\n" );
document.write( "(1 + r) is the growth factor per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "in your problem:<BR>\n" );
document.write( "p would be equal to 375 milligrams.<BR>\n" );
document.write( "r would be -.113 per hour.<BR>\n" );
document.write( "(1 + r) would be (1 - .113) = .887 per hour.<BR>\n" );
document.write( "n would be the number of hours.\r<BR>\n" );
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document.write( "the formula would be simplified to f = 375 * .887 ^ n<BR>\n" );
document.write( "replacing f with D and n with t, the formula would becomes D = 375 * .887 ^ t.<BR>\n" );
document.write( "this would be you solution to this problem.\r<BR>\n" );
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document.write( "if i graph that formula, i would replace D with y and t with x to get y = 375 * .887 ^ x\r<BR>\n" );
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document.write( "here's what the graph would look like.\r<BR>\n" );
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document.write( "the amount of drug after 1 hours would be D = 375 * .887 ^ 1 = 332.675, as shown in the graph.\r<BR>\n" );
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document.write( "the half life of the drug would be calculated as follows:<BR>\n" );
document.write( "D = 375 * .887 ^ t becomes 187.5 = 375 * .887 ^ t.<BR>\n" );
document.write( "divide both sides of the equation by 187.5 to get .5 = .887 ^ t.<BR>\n" );
document.write( "take the log of both sides of the equation to get:<BR>\n" );
document.write( "log(.5) = log(.887 ^ t)<BR>\n" );
document.write( "by log rule that says log(x^t) equals t * log(x), the formula becomes log(.5) = t * log(.887).<BR>\n" );
document.write( "divide both sides of the equation by log(.887) to get log(.5) / log(.887) = t<BR>\n" );
document.write( "solve for t to get t = 5.780547624.<BR>\n" );
document.write( "round to 3 decimal placdes to get t = 5.781<BR>\n" );
document.write( "this agrees with what's on the graph.<BR>\n" );
document.write( "don't forget that D is the same as y on the graph and t is the same as x on the graph.\r<BR>\n" );
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document.write( "i don't know what y=asup(bracket(b),x), stands for.<BR>\n" );
document.write( "a form of the exponential equation would be y = ab^x<BR>\n" );
document.write( "if that's the case, then a would be 375 and b would be .887.<BR>\n" );
document.write( "y = ab^x would then be y = 375 * .887 ^ x<BR>\n" );
document.write( "this is the same as D = 375 * .887 ^ t, with y representing D and x representing t.\r<BR>\n" );
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document.write( "let me know if you have any questions.<BR>\n" );
document.write( "theo </SPAN>\n" );
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