Knowing the half-life of Carbon-14 and the initial mass of Carbon-14 when a tree was cut (a) write an expression that describes the relationship between the remaining mass and time, (b) calculate the remaining mass after $t$ years, and (c) given the remaining mass calculate how many years ago the tree was cut down.
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When a tree was cut down, it contained $\\var{m}$ $\\mathrm{gr}$ of carbon-14.
a) We need to find an expression that connects the number of years and the mass of Carbon-14.
\nWe know that Carbon-14 has a half-life of $5730$ years.We also know that the initial mass of Carbon-14 when the tree was cut down was $\\var{m}$ grams.
\nLet's say it takes $t$ years for the mass Carbon-14 to reach $m$ grams. Since the mass of Carbon-14 halves every $5730$ years, this means that in $t$ years the mass would have halved $\\frac{t}{5730}$ times.
\nSo, our expression would be
\n\\[ \\begin{split} m&=\\var{m}\\times \\left( \\frac{1}{2} \\right) ^\\frac{t}{5730} \\qquad \\text{or} \\qquad m&=\\var{m}\\times0.5^\\frac{t}{5730}\\end{split}\\]
\nb) To find how much Carbon-14 the tree contained after $\\var{t1}$ years, we need to use the expression found in part (a) (either of the suggested forms would work) and set $t=\\var{t1}$:
\n\\[ \\begin{split} m&=\\var{m}\\times 0.5 ^\\frac{\\var{t1}}{5730} \\\\ &=\\var{m}\\times \\simplify{0.5 ^{t1/5730}}\\\\ &=\\var{m}\\times \\simplify{{0.5 ^adv1}} \\\\ &=\\simplify{{m}*{0.5 ^adv1}} \\\\&=\\var{ansb} ~~\\text{[2 d.p.]} \\end{split}\\]
\nTherefore, the tree contained $\\var{ansb}$ grams of Carbon-14 after $\\var{t1}$ years.
\nc) To find how many years ago the tree was cut if it now contains $\\var{m1}$ grams of Carbon-14, we need to use the expression found in part (a) (either of the suggested forms would work), set $m=\\var{m1}$ and then rearrange to calculate:
\n\\[ \\begin{split} \\var{m1}&=\\var{m}\\times 0.5 ^\\frac{t}{5730} \\\\ \\frac{\\var{m1}}{\\var{m}}&=0.5 ^\\frac{t}{5730} \\\\ \\simplify{{m1/m}}&=0.5 ^\\frac{t}{5730} \\end{split}\\]
\nTaking $\\log_{0.5}( )$ of both sides:
\n\\[ \\begin{split} \\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=\\frac{t}{5730} \\\\ 5730 \\times\\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=t \\\\ \\var{adv2}&=t \\end{split}\\]
\nTherefore, the tree was cut approximately $\\var{ansc}$ years ago.
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\n$m=$[[0]] $\\mathrm{gr}$
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\n[[0]]
\nGive your answer in grams to 2 decimal places if needed.
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\n[[0]]
\nGive your answer to the nearest year, if needed.
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