document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1190364>Question 1190364</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Carbon-14 is used to determine the age of artifacts. It has a half-life of 5700 years. This means that<BR>\n" );
document.write( "in 0 years 100%, of the original amount of carbon-14 remains in the artifact. In 5700 years 50% of<BR>\n" );
document.write( "the original amount of carbon-14 remains.<BR>\n" );
document.write( "a) Suppose that 40% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.<BR>\n" );
document.write( "b) Suppose that 75% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.<BR>\n" );
document.write( "c) Suppose that 2% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.<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 #821986 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=821986\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the half life formula is:<BR>\n" );
document.write( ".5 = (1 + r) ^ 5700<BR>\n" );
document.write( "1 + r is the growth factor per year.<BR>\n" );
document.write( "take the 5700th root of both sides of this equation to get:<BR>\n" );
document.write( ".5 ^ (1/5700) = 1 + r<BR>\n" );
document.write( "solve for 1 + r to get:<BR>\n" );
document.write( "1 + r = .9998784026.<BR>\n" );
document.write( "that number is rounded to the number of decimal digits displayed in the calculator.<BR>\n" );
document.write( "to get the most accurate value when using that factor, use the internally stored value rather than the displayed value.<BR>\n" );
document.write( "to confirm the value is correct, then:<BR>\n" );
document.write( "1 * .9998784026 ^ 5700 = .4999999286, using the displayed value.<BR>\n" );
document.write( "1 * .999878402... ^ 5700 = .5, using the stored value in the calculator,.\r<BR>\n" );
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document.write( "now that you know the rate per year, you can solve for the rest of the problem.\r<BR>\n" );
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document.write( "a) Suppose that 40% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.<BR>\n" );
document.write( "b) Suppose that 75% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.<BR>\n" );
document.write( "c) Suppose that 2% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.\r<BR>\n" );
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document.write( "start with .5 = x ^ y<BR>\n" );
document.write( "x is equal to our stored value of .9998784026.<BR>\n" );
document.write( "note that this is the displayed value of x.<BR>\n" );
document.write( "the actual value of x is stored in the calculator and has more decimal digits than the displayed value.\r<BR>\n" );
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document.write( "we want to solve for y.<BR>\n" );
document.write( "take the log of both sides of this equation to get:<BR>\n" );
document.write( "log(.5) = y * log(x).<BR>\n" );
document.write( "divide both sides of this equaiton by log(x) to get:<BR>\n" );
document.write( "log(.5) / log(x) = y<BR>\n" );
document.write( "solve for y to get:<BR>\n" );
document.write( "y = 5700.<BR>\n" );
document.write( "note that, since x = .9998784026, the equation is really:<BR>\n" );
document.write( "log(.5) / log(.998784026) = y.<BR>\n" );
document.write( "once again, you are using the stored value of x and not the displayed value of x, for greater accuracy.\r<BR>\n" );
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document.write( "this confirms that we can find the value of y using this formula.<BR>\n" );
document.write( "the value of y is the number of years.\r<BR>\n" );
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document.write( "now to solve the rest of the problems.\r<BR>\n" );
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document.write( "a) Suppose that 40% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.\r<BR>\n" );
document.write( "<BR>\n" );
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document.write( "if 40% of the original carbon remains, then the formula becomes:<BR>\n" );
document.write( "log(.4) / log(x) = y<BR>\n" );
document.write( "solve for y to get:<BR>\n" );
document.write( "y = 7534.990141.\r<BR>\n" );
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document.write( "b) Suppose that 75% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if 75% of the original carbon remains, then the formula becomes:<BR>\n" );
document.write( "log(.75) / log(x) = y<BR>\n" );
document.write( "solve for y to get:<BR>\n" );
document.write( "y = 2365.713746.\r<BR>\n" );
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document.write( "c) Suppose that 2% of the original carbon remains in a piece of pottery.<BR>\n" );
document.write( "Estimate the age of the pottery.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if 2% of the original carbon remains, then the formula becomes:<BR>\n" );
document.write( "log(.02) / log(x) = y<BR>\n" );
document.write( "solve for y to get:<BR>\n" );
document.write( "y = 32169.98028\r<BR>\n" );
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document.write( "the equation can be graphed as shown below:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2022/020303.jpg\" >\r<BR>\n" );
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document.write( "let me know if you have any questions.\r<BR>\n" );
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document.write( "theo\r<BR>\n" );
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document.write( " </SPAN>\n" );
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