document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1146030>Question 1146030</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Calc 2 Problem:\r<BR>\n" );
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document.write( "The rate at which sugar dissolves in water is proportional to the amount that remains. Suppose that 10 lb of sugar is placed in a container of water at 1:00pm and one half is dissolved at 4:00pm.\r<BR>\n" );
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document.write( "a) Find a formula for the amount of sugar remaining in the container after t hours. How long will it take to have 3 lb of sugar remaining?\r<BR>\n" );
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document.write( "b) How much of the 10 lb will remain in the container at 8:00pm? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #767334 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=767334\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a formula that can be used is 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 rate per time period<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "in this problem:<BR>\n" );
document.write( "f = 5<BR>\n" );
document.write( "p = 10<BR>\n" );
document.write( "r = what you want to find<BR>\n" );
document.write( "n = 3 hours\r<BR>\n" );
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document.write( "formula becomes 5 = 10 * (1 + r) ^ 3<BR>\n" );
document.write( "divide both sides of this formula by 10 to get .5 = (1 + r) ^ 3<BR>\n" );
document.write( "take the third root of both sides of this equation to get .5 ^ (1/3) = 1 + r<BR>\n" );
document.write( "subtract 1 from both sides of this equation to get .5 ^ (1/3 - 1 = r<BR>\n" );
document.write( "solve for r to get r = -.206299474.\r<BR>\n" );
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document.write( "replace r in the original equation with that to get:\r<BR>\n" );
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document.write( ".5 = (1 - .206299474) ^ 3 = .5<BR>\n" );
document.write( "this confirms the value of r is correct.\r<BR>\n" );
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document.write( "in t hours, when r = -.206299474 and p = 10 and f = 3, the formula of f = p * (1 + r) ^ n becomes:\r<BR>\n" );
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document.write( "3 = 10 * (1 - .206299474) ^ t<BR>\n" );
document.write( "divide both sides by 10 to get:<BR>\n" );
document.write( "3/10 = (1 - .206299474) ^ t<BR>\n" );
document.write( "take the log of both sides to get:<BR>\n" );
document.write( "log(3/10) = log(1 - .206299474) ^ t)<BR>\n" );
document.write( "by log rules, this becomes:<BR>\n" );
document.write( "log(3/10) = t * log(1 - .206299474)<BR>\n" );
document.write( "divide both sides by log(1 - .206299474) to get:<BR>\n" );
document.write( "log(3/10) / log(1 - .206299474) = t<BR>\n" );
document.write( "solve for t to get:<BR>\n" );
document.write( "t = 5.2108967682\r<BR>\n" );
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document.write( "replce t in the original equation to get:<BR>\n" );
document.write( "3 = 10 * (1 - .206299474) ^ 5.2108967683 =   3<BR>\n" );
document.write( "this confirms the number of time periods is correct.<BR>\n" );
document.write( "the 10 pounds of undissolved sugar will be reduced to 3 in that number of hours.\r<BR>\n" );
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document.write( "if the sugar is placed in the container at 1:00 pm, then at 8:00 pm, the number of hours it has been in the container is 7 hours.\r<BR>\n" );
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document.write( "the formula becomes f = 10 * (1 - .06299474) ^ 7<BR>\n" );
document.write( "solve for f to get:<BR>\n" );
document.write( "f = 1.984251315<BR>\n" );
document.write( "that's how much sugar has not been dissolved in 7 hours.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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